KINEMATICS SESSION 1& 2 AIM
Basic definitions and their explanations
BASIC DEFINITIONS AND THEIR EXPLANATIONS
Distance: The actual path length of the particle from initial to final positions of journey.
Symbol: “S” SI unit: m The distance travelled by the particle is independent of direction of motion and it is scalar quantity.
Displacement: The shortest distance between initial and final
positions of journey.
Displacement is a vector quantity and its magnitude is the shortest distance between the initial and final positions of journey.
Displacement and distance can be evaluated not only between initial and final positions but also between any two points of journey.
Consider A and B are two positions of particle in its journey.
A = = = y
→Final position vector → Initial position vector → Displacement vector
Hence displacement is “final position vector” minus “initial position vector”. Displacement can have zero magnitude, but distance
travelled can never be zero.
If r =a1 ̂+b1 ̂+c1 and r = a2 ̂+b2 ̂+c2 then r = r − r =
(a2 ̂+b2 ̂+c2 ) – (a1 ̂+b1 ̂+c1 )
= (a2–a1) ̂+ (b2–b1) ̂+ (c2–c1)
Speed: The distance travelled per unit time Symbol: “v” SI unit: ms-1
Speed is the scalar quantity
Velocity: The displacement of the particle per unit time. Symbol: “v”: SI unit: ms-1
Velocity is a vector quantity; its direction is same as that of displacement.
Let us consider an example to discuss their concepts:
A B
Let a particle move from point A to point B as shown:
Along path 1, distance travelled is 100m, time taken is 10s Along path 2, distance is 200m, time taken is 15s
Along path 3, distance travelled is 150m, time taken is 12s
2
1
Along path 1: distance =s=100m:speed = ms–1 = 10ms–1;
Velocity = ̅ = ms–1 (from A to B) ⟹ ̅ = 10ms–1 from A to B
(displacement is same distance along path 1)
Along path 2: displacement = 100 m; time = 15 s speed = = ms–1;
Velocity = ̅ = = ms–1 (from A to B)
Along path 3: displacement = 100 m; distance = 150m; time = 12s
Speed = = ms–1; velocity = = ms–1 (from A to B)
Note: The direction of displacement and velocity is same for all the three paths i.e. imagine a vector drawn from A to B. The direction of vector gives the direction of displacement and velocity.
Average Speed: - When any object travels with non-uniform speeds during different time intervals or travelling different segments of path, the ratio of total distance covered to the total time taken is called 'Average Speed'.
e.g. Average Speed= ( )
( )
If an object travels with speeds v1, v2, v3...vN, during time intervals t1, t2, t3 ...tN, travelling for distances s1, s2, s3...sN, its
average speed can be discussed in following three ways.
When Distances & Speeds are When Distances & Time intervals are When Speeds & Mentioned Time intervals are mentioned
Average Speed ( ̅) = …. …. Average Speed ( ̅) = + … . Average Speed ( ̅) = … … If s1 = s2 = s3... =sN= then If t1 = t2 = t3...tn= then Average Speed ( ̅) = … Average Speed ( ̅) = …
i.e. Harmonic Mean of Velocities i.e. Arithmetic Mean of Velocities These are Basic independent cases. In practice you may have combination of two or more.
Average velocity: It is defined as the ratio of displacement to time taken by the body
Average Speed ( ̅) =
; ⃗ = ⃗
Instantaneous velocity:
The term “instantaneous” is used to explain an event happening in a very smallinterval of time. Usually instantaneous time is denoted as “dt”. The numerical value of ‘dt” is nota definite constant, but we can say that it less than one and close to zero i.e. about 0.01s or 0.001s. (Let us not assign a fixed value for it. Its depends on the context of explanation).
Let us consider a situation of a particle moving along a straight line from A to B. At a time t = 0 (usually called initial time) the particle is at A and at a time t = 10s. Let AB = 100m. As the particle crosses the point C, can we guess time taken to cross this point. The answer is simple. “Impossible”. If we consider two nearest points, one point D just before reaching C and another point E just after crossing C. The
time interval from D to E is taken as instantaneous time i.e. “dt”. The displacement/ distance between D and E are denoted as / i.e.
is called infinitesimal (very small) displacement ( ) DE is called infinitesimal distance (ds)
Instantaneous speed: It is the infinitesimal distance travelled per infinitesimal time.
Vinst =
Instantaneous velocity: It is the ratio of infinitesimal displacement to infinitesimal time.
in st =
Note: Always it is assumed that the particle is along a straight line path in this infinitesimal time.
Uniform speed: Speed of the particle is said to be uniform, if it covers equal distances in equal intervals of time taken, however small the intervals are.
i.e. VAB = VAc = VCB = VDE
Non-uniform Speed: Speed of the particle is said to be non-uniform,
if it covers unequal distances in equal intervals of time, however small the intervals are.
Non-uniform velocity: Velocity of the particle is said to be non-uniform, if it travels unequal displacements in equal intervals of time, however small the intervals are.
Uniform speed is possible for any path of particle i.e. straight line, curved paths. Uniform velocity is possible only for a straight line path, because velocity is a vector quantity i.e. for velocity to be constant, the direction of velocity should be same.
CLASS EXERCISE :
1] A circular park has a radius of 1 km. A man standing from the center of park, walking towards East and going along the circumference of the park reaches the North gate. What is his displacement?
a)√2 km b) 1.57 km c) 1 km d) 2.57 km
2] If a cyclist takes one minute to complete half revolution on a circular path 120m radius, what is the average velocity?
a) 1 m/s b) 2 m/s c) 3 m/s d) 4 m/s
3] A car covers the 1st half of the distance between two places at a speed of 40 km/hr and the second half at 60 km/hr. The average speed of the car is
a) 100 km/hr b) 48 km/hr c) 50 km/hr d) 25 km/hr
4] An aeroplane moves 400m towards north, 300m towards west and then 1200m vertically upwards. Then its displacement from the initial position is
a) 1300m b) 1400m c) 1500m d) 1600m
5] An ant starts from one corner of a cube of side length 3m and reaches the diagonally opposite corner. The displacement is
SESSION-3, 4 & 5 AIM
Definition of acceleration, explanation of equations of motion. Definition of acceleration, explanation of equations of motion.
Acceleration: The rate of change of velocity is acceleration.
The direction of acceleration is the direction of change in velocity There is misconception that direction of acceleration is either along the direction of motion or opposite to direction of motion. It need not be parallel to direction of motion.
= ; 1 and 2 need not be in the same direction 2– 1need not be either in the direction of 1 or 2.
For a particle moving along a straight line, direction of acceleration is either in the direction of motion (in case of increasing velocity) or in the opposite to direction of motion (in case of decreasing velocity)
Uniform acceleration: The body is said possess uniform
acceleration if it executes equal change in velocity in equal intervals of time, however small the intervals are.
There are two possible paths for this type of motion (i) straight line (ii) parabolic
Non-uniform acceleration: The body is said to possess
non-uniform acceleration, if it executes unequal change in velocity in equal intervals of time, however small the intervals are.
Average Acceleration: - If Velocity of an object changes from to during an interval of t=t1 to t=tz, the Average Acceleration is given as
⃑ Average=< ⃑>= =
⃑ Instantaneous = ⃑ = lim → =
You know that the definitions might be boring but simple but applications are always interesting. Therefore, let's come to application
For Motion in One Dimension you have already learnt Equations of Motion in One Dimension
v=u+at s = ut + at2 vz = uz + 2as
We can not only establish same set of equation without graphs but we can derive any result if we use Calculus Technique.
Calculus Techniques:- The Equations of Motion in One Dimension are consistent to constant acceleration only. Therefore to start with consider constant acceleration.
i.e. ⃑ = constant t Therefore by definition For 2nd Equation we can use ⃑ =
∴ d ⃑ = ⃑dt⟹ ∫ ⃑ = ∫ ⃑. ⟹[ ⃑] = ⃑[ ] ⟹ ⃑ − ⃑ = ⃑ = ⃑( − 0) ⟹ ⃑ = ⃑ +
Therefore you got your 1st Equation ⃑ = ⃑ + ⃑t or v = u + at because for Motion in One Dimension only “+ve” & “-ve” sign are more than sufficient to rep represent direction. "+ve" means away from Origin i.e. towards Right or Upwards and "-ve" means towards the Origin i.e. towards left or Downwards. ⃑ = ⃑ + ⃑t Sufficient for any type of motion with Constant Acceleration.
For 2nd Equation use ⃑ = ⃑
∴ d ⃑ = ⃑ ⟹ ∫ ⃑ ⟹ ∫⃑⃑[ ⃑ = ∫ ( ⃑ + ⃑ ). ⟹ [ ⃑]⃑⃑ = ⃑. + ⃑.
⟹ ⃑ − ⃑ = ⃑. + ⃑. Where ⃑ − ⃑ = ⃑ (Displacement)
Therefore you got your 1st Equation ⃑= ⃑ + ⃑ = v = u + at because for Motion in One Dimension only "+ve" &" -ve "sign are more than sufficient to represent direction. ”+ve" means away from Origin i.e. towards Right or Upwards and "-ve" means towards the Origin i.e. towards left or Downwards.
NOTE: ⃑ = ⃑ + ⃑ is sufficient for any type of motion with Constant Acceleration.
For 2nd Equation use ⃑ = ⃑
∴ d ⃑= ⃑. ⟹ ∫⃑⃑ ⃑ = ∫ ⃑ ⟹ ⃑ = ∫ ( ⃑ + ⃑ ) ⟹ [ ⃑]⃑⃑ = ⃑. + ⃑.
⟹ ⃑ − ⃑ = ⃑. + ⃑. Where ⃑ − ⃑ = ⃑ (Displacement)
Therefore you got your 2nd Equation. ⃑ = ⃑. + ⃑. or s = u.t. +
a.t2 because for Motion in One Dimension only "+ve" &”-ve" sign are more than sufficient to represent direction. "+ve" means away from Origin i.e. towards Right or Upwards and "- ve" means towards the Origin i.e. towards left or Downwards.
Note: ⃑ = ⃑. + ⃑. is sufficient for any type of motion with Constant Accelection.
For 3rd Equation method is little bit different. As we know by definition ⃑ = ⃑ but we require Velocity and Displacement form. Therefore ⃑ = ⃑× ⃑ ⟹ ⃑ = ⃑. ⃑ ⃑. ⃑ ⟹ ⃑.d ⃑ =∫ ⃑ ⃑ = ⃑ ∫⃑ ⃑⃑ ⃑⟹ ⃑ ⃑ ⃑ = ⃑. [ ⃑]⃑⃑ ⟹ ⃑ ⃑ =2. ⃑( ⃑ − ⃑ ) Where ⃑ − ⃑ = ⃑ (Displacement)
Therefore you got your 3rd Equation. ⃑ = + 2. ⃑. ⃑or ⃑ = + 2. . because for Motion in One Dimension only "+ve" & "-ve" sign are more than sufficient to represent direction. "+ve" means away from Origin i.e. towards Right or Upwards and " - ve" means towards the Origin i.e. towards left or Downwards.
NOTE: ⃑ = + 2. ⃑. ⃑ is sufficient for any type of motion with Constant Acceleration.
Instantaneous Acceleration: Instantaneous Acceleration is the Acceleration of the object at a particular instant. Here we get to use Calculus once more.
Sn = u + (2n - 1); Sn is displacement of the particle in the n-th second
The term n-th second is to be understood with clarity. This instant with duration of 1 second.
The instant with duration of ‘t’ seconds can have duration less than or greater than or equal to one second. But the n-th second always has the interval “one second”.
The last equation should not be used to calculate displacement if the interval of time not equal to 1 second.
Example:
(i) A car travels for six seconds. The last second of travel is 6th second. (Interval of last second is one second)
(ii) A vertical travel for 6.1 seconds. The last second of travel has an interval of one second. i.e. from 5.1 s to 6.1 s
(iii) A cyclist travels for 0.95 seconds. There is no last second, because the travel time is less than one second.
To calculate displacement from t1 to t2 seconds, use
s = ut + at2 i.e. s1 = ut1 + a and s2 = ut2+ a ⟹(s2-s1) = u (t2 -t1) + a ( - )
To calculate displacement from 5.1s to 6.1s, also use s = ut + at2 along with Sn = u + (2n -1)
To calculate displacement from 5s to 6s. Use Sn = u + (2n -1) with n= 6
For a body moving with uniform acceleration, average velocity is given by and by definition, average velocity =
EXERCISE (A)
1] Moving with uniform acceleration, a body covers 150m during 10sec so that it covers 24 m during the tenth second. Find the initial velocity and acceleration of the body
a) 2 ms–1; 5 ms–2 b) 5ms–1; 2 ms–2 c) 3 ms–1; 4 ms–2 d) 4ms–1;3ms–2
2] A body moving with uniform acceleration covers 100m in the first 10 seconds and 150m in the next 10 seconds. The initial velocity of the body is
a) 15 ms–1 b) 7.5 ms–1 c) 5 ms–1 d) 2.5 ms–1
3] A particle moving with a constant acceleration describes in the last second of its motion th of the whole distance. If it starts
from rest, how long is the particle in motion
a) 5s b) 10s c) 15s d) none
4] A body starts with initial velocity u and moves with uniform acceleration f. If when the velocity has increased to 5u, the acceleration is reversed in direction, the magnitude remaining constant; it turns to the starting point with velocity of
a) – u b) –6u c) –7u d) –9u
5] A body moving with uniform acceleration travels a distance Sn = (0.4n + 9.8) m in nth sec. Find the initial velocity of the body in ms–1
6] The displacement of particle is zero at t = 0 and it is x at t = t. It starts moving in the positive x direction with a velocity which varies as V = K, where K is constant. The relation between V and t is
a) Kt/2 b) Kt3/2 c) K2t/2 d) 2Kt2
7] A particle starts from rest with constant acceleration for 20sec. If it travels a distance y1 in the first 10 sec and a distance y2 in the next 10 sec then
a) y2 =2y1 b) y2 =3y1 c) y2 =4y1 d) y2 =5y1
8] The ends of a moving train with a constant acceleration pass a certain point with velocities u and v. What is the velocity V with which the mid–point of the train passes through the same point?
a) v = b) v = √ +
c) v = d) V =
9] The velocity of a body at an instant is 20 ms–1. After 5s the velocity is 30 ms–1. How many seconds earlier from the instant, it might have started? Assume acceleration is uniform a) 8sec b) 10sec c) 6sec d) 7sec
10] The engine of a train moving with uniform acceleration passes an electric pole with a velocity u and the last compartment with a velocity V, the length of train that has passed the pole when the velocity is , if the total length of the train is l
a) ( )
( ) b) c) d)
( )
SESSION - 6 AIM
Motion with Variable Acceleration Motion with Variable Acceleration
(i) If acceleration is a function of time then a = f (t) then v = u +∫ ( ) and s = ut +∫(∫ ( ) )
(ii) If acceleration is a function of distance a = f(x) then v2=u2+2∫ ( )
(iii) If acceleration is a function of velocity a = f (v) then t = ∫
( ) and x = x0 + ∫
( )
ILLUSTRATION:
Problem1. An electron starting from rest has a velocity that increases linearly with the time that is v = kt where k = 2 m/sec2. The distance travelled in the first 3 seconds will be
a) 9 m b) 16 m c) 27 m d) 36 m Sol: (a) x = ∫ = ∫ 2 = 2 = 9m
Problem2.The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity The distance travelled by the particle in time t will be a) V0t + bt2 b) V0t + bt3 c) V0t + bt3 d) V0t + bt2
Sol: (c) ∫ =∫ = ∫ ( )
⟹ v2 – v1= = + = +
S = ∫ + ∫ = +
Problem3. The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 seconds
a) 4 b)12 c)6 d) 8
Sol: d) u = at⟹ =at
S= ∫ = a =8a
Problem4. A body A moves with a uniform acceleration and zero initial velocity. Another body B starts from the same point moves in the same direction with a constant velocity v. The two bodies meet after a time t. The value of t is
a) b) c) d)
Sol: (a) Let they meet after time’t’. Distance covered by body A Distance covered by body B= vt and = ∴ =
Problem5. A student is standing at a distance of 50metres from the bus. As soon as the bus starts its motion with an acceleration of 1ms-2, the student starts running towards the bus with a uniform velocity Assuming the motion to be along a straight road, the minimum value of , so that the students is able to catch the bus is
a) 5 ms-1 b)8 ms-1 c) 10 ms-1 d)12 ms-1
Sol: (c) Let student will catch the bus after sec. So it will cover distance ut. Similarly distance travelled by the bus will be for the given condition ut = 50 + =50 + = + (As a = 1 m/s2)
Problem6. A particle is moving eastwards with velocity of 5 m/s. In 10 sec the velocity changes to 5 m/s northwards. The average acceleration in this time is
a) Zero b) toward north-west c) √ m/s2toward north-east d) m/s2 toward north-west Sol: b)Δ ⃑ = ⃑ – ⃑ Δ = + − 2 90 = 5 + 5 = 5√2 1 v s m / 5 2 90o s m / 5 1
Δ = 5√2 Average acceleration = = √ =
√ / toward north-west (As clear from the figure).
Problem7. A body starts from the origin and moves along the x-axis such that velocity at any instant is given by where t is in second and velocity is in m/s. What is the acceleration of the particle, when it is 2m from the origin?
a) 28 m/s2 b) 22 m/s2 c) 12 m/s2 d) 10 m/s2
Sol: (b) Given that = 4 − 2
x = ∫ at x = t4-t2+c at t =0, x=0 ⟹C=0
When particle is 2m away from the origin
2= t4–t2 ⟹t4–t2–2=0 ⟹ (t2–2) (t2+1) = C ⟹ t=√2 sec
a = = (4 − 2 ) = 12 − 2 ⟹ = 12 − 2for t = √2 sec ⟹a = 12×(√2)2–2⟹ =22 m/s2
Problem 8. A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 sec on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is
a) 3 m/s2 b) –3 m/s2 c) 0.3 m/s2 d) –0.3 m/s2
Sol: (b) Let particle moves towards east and by the application of constant force it moves towards west
⃑ = +10 / dx and ⃑ = −2 / Acceleration = = ⃑ ⃑ ⟹ = (−2) − (10) 4 = − 12 4 = −3 /
Problem9. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is
a) 6m b) 12m c) 18m d) 24m
Sol (d): v2 = u2 - 2as ⟹ s = ⟹s u2 (As a = constant) = ⟹ s2 = 4s1 = 4 x 12 = 24m
Problem10. The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be
a) 10 x 104 m/s2 b) 12 x 104 m/s2 c) 13.5 x 104 m/s2 d) 15 x 104 m/s2
Sol: (d) u = 200 m/s, v = 100 m/s, s = 0.1 m;
a = = ( ) ( )
× . = 15 x 104 m/s2
Problem 11.A body A starts from rest with an acceleration. After 2 seconds, another body B starts from rest with an acceleration. If they travel equal distances in the 5th second, after the start of A, then the ratio is equal to
a) 5 : 9 b) 5 : 7 c) 9 : 5 d)9 : 7
Sol: (a) By using Sn = u+ (2n -1), Distance travelled by body A in 5th second= 0 + (2 x 5 -1)
Distance travelled by body B in 3rd second is = 0+ (2×3–1) According to problem: 0+ (2×5–1) = 0+ (2×3–1)⟹ 9 a1 = 5a2 ⟹ =
Problem12. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms-1. If the change in velocity of the body is 0.18ms-1 during this time, its uniform acceleration is a) 0.01 ms-2 b) 0.02 ms-2 c) 0.03 ms-2 d) 0.04 ms-2 Sol: (b) Time = = . . =9sec and Acceleration = = . =0.02 m/s2.
Problem13. A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec
a) 8.3 m b) 9.3 m c) 10.3 m d) None of above Sol: a) Let initial (t = 0) velocity of particle = u for first 5sec of
motion s5 = 10 metre, so by using s= ut + at2
10 = 5u + a (5)2⟹ 2u + 5a = 4 ... (i) for first 8 sec of motion s8 = 20 metre
20 = 8u + a (8)2⟹ 2u + 8a = 5 ... (ii) By solving (i) and (ii) u = m/s a = m/s2
Now distance travelled by particle in total 10 sec. s10 = u×10+ a(10)2
by substituting the value of u and a we will get s10 = 28.3 m So the distance in last 2 sec = s10 - s8 = 28.3 - 20 = 8.3 m
Problem14. A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S1, S2 and S3 in the first five seconds, second five seconds and next five seconds respectively the relation between S1, S2 and S3 is
a) S1 = S2 = S3 b) 5S1 = 3S2 = S3 c) S1 = S2 = d) S1 = S2 = S3
Sol: (c) Since the body starts from rest. Therefore u = 0 S1 = a(5)2 =
S1 + S2 = a (10)2 = ⟹ S2 = -S1 = 75 S1 + S2 + S3 = a (15)2 = ⟹S3 = -S2 - S1 = Thus Clearly S1 = S2 = S3
Problem15. If a body having initial velocity zero is moving with uniform acceleration the distance travelled by it in fifth second will be
a) 36 metres b) 40 metres c) 100 metres d) Zero Sol: (a) sn= u + a (2n - 1) = 0 + (8) [2 x 5 - 1] = 36 metres
Problem16. The engine of car produces acceleration 4m/sec2 in the car, if this car pulls another car of same mass, what will be the acceleration produced
a) 8 m/s2 b) 2 m/s2 c) 4 m/s2 d) m/s2
Sol: (b) F = ma a if F = constant. Since the force is same and the effective mass of system becomes double = = a2=
Problem17. A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second?
a) 7/5 b) 5/7 c)7/3 d)3/7
Sol: (a) As Sn (2 − 1), = CLASS EXERCISE :
1] If the relation between distance x and time t is of the from t = x2+ x here and being appropriate constants, then the retardation of the particle is
a) 2 v3 b) 2 v3 c) 2 v3 d)2 2v3
2] A particle moves along a straight line according to the law = + 2 + . . The acceleration of the particle varies as a) S–3 b) S2/3 c) S3 d) S5/2
3] A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as = √ , where k is a positive constant. At the initial moment the velocity of the point is equal to v0. What distance will it traverse before it stops? What time will it take to cover that distance?
4] If the displacement of the particle varies with time as x1/2=t+7, then
a) velocity of the particle is inversely proportional to t b) velocity of the particle is proportional to t
c) velocity of the particle is inversely proportional t1/2. d) the particle moves with a constant acceleration.
5] A point moves in a straight line under the retardation kv2. If the initial velocity is u, the distance covered in t second is
a) kut b) log (k ut)
6] A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a force F(x) = – (k/2x2) where k = 10–2 Nm2. At time t = 0 it is at x = 1 m and its velocity v = 0. (i) Find its velocity when it reaches x = 0.5 m
7] A particle moving in a straight line has an acceleration of (3t - 4) ms-2 at timet seconds. The particle is initially 1m from O, a fixed point on the line, with a velocity of 2 ms-1. Find the times when the velocity is zero. Find also the displacement of the particle from O whent = 3.
SESSION 7, 8 & 9 AIM
Graphical Representation of motion Motion of Body under Gravity (Free Fall).
The force of attraction of earth on bodies is called force of gravity. Acceleration produced in the body by the force of gravity, is called acceleration due to gravity. It is represented by the symbol g.
In the absence of air resistance, it is found that all bodies (irrespective of the size, weight or composition) fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude (h << R) is called free fall.
An ideal one-dimensional motion under gravity in which air resistance and the small changes in acceleration with height are neglected.
(1) If a body dropped from some height (initial velocity zero)
(i) Equation of motion: Taking initial position as origin and direction of motion (i.e., downward direction) as a positive, here we have
u = 0 v h g v g h t 2 gh v 2 g v h 2 2
u = 0 [As body starts from rest]
a = +g [As acceleration is in the direction of motion]
v = g t … (i)
h = .... (ii)
=2gh .... (iii)
hn= (2n+1) .... (iv)
(ii) Graph of distance velocity and acceleration with respect to time:
(iii) As h = (1/2)gt2, i.e., h t2, distance covered in time t, 2t, 3t, etc., will be in the ratio of 12 : 22 : 32, i.e., square of integers.
(iv) The distance covered in the nth sec, hn = (2 − 1)So
distance covered in I, II, III sec, etc., will be in the ratio of 1: 3: 5, i.e., odd integers only.
(2) If a body is projected vertically downward with some initial velocity
Equation of motion: h = ut + gt2
= u2+2gh hn= u+ (2n –1)
(3) If a body is projected vertically upward
(i) Equation of motion: Taking initial position as origin and direction of motion (i.e., vertically up) as positive
a =–g [As acceleration is downwards while motion upwards]
s t a g t v t tan = g
So, if the body is projected with velocity u and after time t it reaches up to height h then
= − h = ut– gt2 = u2–2gh hn= u – (2n –1)
(ii) For maximum height v = 0 So from above equation u = gt, h = gt2 and = 2 ℎ
(iii) Graph of distance, velocity and acceleration with respect to time (for maximum height) :
It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of air resistance, all bodies fall on the surface of the earth with the same rate.
(4) In case of motion under gravity for a given body, mass, acceleration, and mechanical energy remain constant while speed, velocity, momentum, kinetic energy and potential energy change. g u g h t 2 gh u 2 g u h 2 2 u v = 0 h s t (u/g) (u2/2g) (u/g) (2u/g) t v O – v + – t a O g + – a
(5) The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity i.e., t =
(2ℎ/ )and v = 2 ℎ
(6) In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance. Time of descent (t1) = time of ascent (t2) = u/g
Total time of flight T = t1 + t2 =
(7) In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection. As well as the magnitude of velocity at any point on the path is same whether the body is moving in upwards or downward direction.
(8) A ball is dropped from a building of height h and it reaches after t seconds on earth. From the same building if two ball are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1 and t2 seconds respectively then
t = √
u u u=0
(9) A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than the time of descent. t2> t1
Let u is the initial velocity of body then time of ascent = and h =
( ) where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will work vertically downward.
For downward motion a and g will work in opposite direction because a always work in direction opposite to motion and g always work vertically downward.
So h = (g-a) ⟹ ( ) = ( − ) ⟹ t2=
( )( )
Comparing t1 and t2 we can say that t2> t1 since (g + a) > (g - a)
(10) A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distance of 1m each will then be in the ratio of the difference in the square roots of the integer’s i.e.
√1,(2–√1),(√1–√1)…….(√4 − √3),……. u = 0 1 1 t 1 2 2 t 2 3 3 t 3 4 4 t 1 m 1 m 1 m 1 m
CLASS EXERCISE :
1] The variation of velocity with time of a particle moving along a straight line is illustrated in the following figure. The distance traveled by the particle in 4sec. is
a) 60m b) 45m c) 55m d) 50m
2] In the given v-t graph, the distance travelled by the body in 5 second will be
a) 20 m b) 40 m c) 80 m d) 100 m
3] For the displacement- time graph shown in fig. the ratio of the magnitudes of the speeds during the first two second and the next four second is
a) 1: 1 b) 2 : 1 c) 1 : 2 d) 3 : 2 m /s 20 10 A 4 3 2 1 30 E B C D 20m Time 0s 2s 6s d is p la ce m en t
4] The velocity- time graph of a particle moving along a straight line is shown in fig. The displacement of the body in 5 second is
a) 2.5m b) 1m c) 2m d) 3m V (m /s ) -1 1 0 4 3 5 1 2 t(ins) 1 2 2 1.5 1.5
SESSION 10, 11 & 12 AIM
Motion of a body under gravity
Projectile -Motion in Two Dimensions
Motion in a plane is called as motion in two dimensions e.g., projectile motion, circular motion etc. For the analysis of such motion our reference will be made of an origin and two co-ordinate axes x and y. Position of particle is known by knowing its co-ordinate (x, y). Velocity of particle will be resultant of velocities in x and y direction Vx and Vy. Similarly acceleration will be in the two directions. For analysis of such motion we analyses the motion along two axes independently i.e., while dealing motion in x-direction we need not to think what is going on in y-x-direction and vice versa.
We have to study about projectile motion, circular motion and relative velocity under the head of motion in two dimensions.
Projectile Motion
A projectile is a particle, which is given an initial velocity, and then moves under the action of its weight alone. If the initial velocity is vertical, the particle moves in a straight line and such motion we had already discussed in ‘motion in one dimension as motion under gravity’. Here we are going to discuss the motion of particle which is projected obliquely near the earth’s surface. While discussing such motion we shall suppose the motion to be within such a moderate distance from the earth’s surface, that acceleration due to gravity may be considered to remain sensibly constant. We shall
also neglect the resistance of air and consider the motion to be in vacuum.
Important Terms used in projectile motion
When a particle is projected into air, the angle that the direction of projection makes with horizontal plane through the point of projection is called the angle of projection; the path, which the particle describes, is called the trajectory, the distance between the point of projection and the point where the path meets any plane draws through the point of projection is its range; the time that elapses in air is called as time of flight and the maximum distance above the plane during its motion is called as maximum height attained by the projectile.
Analytical treatment of projectile motion
Consider a particle projected with a velocity u of an angle with the horizontal from earth’s surface. If the earth did not attract a particle to itself, the particle would describe a straight line; on account of attraction of earth, however, the particle describes a curved path. This curve will be proved later to be always a parabola.
Let us take origin at the point of projection and x-axis and y-axis along the surface of earth and perpendicular to it respectively as shown in figure.
By the principle of physical independence of forces, the weight of the body only has effect on the motion in vertical direction. It,
y O P u x O
therefore, has no effect on the velocity of the body in the horizontal direction, and horizontal velocity therefore remains unaltered.
Motion in x- direction:
Motion in x - direction is motion with uniform velocity. At t = 0, x0 = 0 and ux = u cos
Position after time t, x = x0 + uxt ⟹ x = (u cos ) t
Velocity of any time t, vx = ux ⟹vx= u cos
Motion in y-direction:
Motion in y-direction is motion with uniformly acceleration When, t = 0, y0 = 0, uy = u sin and ay = -g
∴ After time‘t’, vy = uy + ay t ⟹vy = u sin - gt
y = y0 + uyt + ayt2 ⟹y = uyt + ayt2 y = (u sin ) t
-Also, = + 2ayy ⟹ = u2 sin2 - 2 gy
Time of Flight (T):
Time of flight is the time during which particle moves from O to O i.e., when t = T, y = 0
O = u sin T- gT2 T = Range of projectile (R):
Range is horizontal distance travelled in time T. i.e., R = x (in time T)
∴ From equation (ii) R = ucos .T
= ucos R =
Maximum height reached (H):
At the time particle reaches its maximum height velocity of particle becomes parallel to horizontal direction i.e., vy = 0 when y = H
∴ From equation (v)
0 = u2 sin2 − 2 gH; H =
Equation of trajectory:
The path traced by a particle in motion is called trajectory and it can be known by knowing the relation between x and y.
From equation (1) and (2) eliminating time t we get y = x tan - sec2
This is trajectory of path and is equation of parabola. So we can say the path of particle is parabolic.
Velocity and direction of motion after a given time: After time‘t’ vx = ucos and vy = usin -gt
= + ( − )
If direction of motion makes an angle with horizontal
tan = = ⟹ =
Velocity and direction of motion at a given height:
At a height ‘h’, Vx = ucos
And Vy = − 2 ℎ
Resultant velocity v = + ; v = − 2 ℎ
Note that this is the velocity that a particle would have at height h if it is projected vertically from ground with u.
SOME IMPORTATN POINTS REGARDING PROJECTILE MOTION OVER A HORIZONTAL PLANE
i) For a given velocity of projection, the range of horizontal plane will be maximum when angle of projection is 450.
We have range of projectile. R =
Therefore if we keep on increasing range will increase and then decrease. Its value will be maximum when sin2 is maximum i.e., = 450
Also, maximum range Rmax =
ii) For a given range and given initial speed of projection, there are two possible angle of projection which are complementary angle i.e., if one is other will be (900- ).
ClASS EXERCISE (A) :
1] In the last second of free fall, a body covered 3/4 of its total path. Then the height from which the body is released will be a) 4.9m b) 9.8m c) 19.6m d) 39.2m
2] A freely falling body travels a distance X in the nth second. In the next second if it travels a distance, Y. Then
a) X+Y=g b) X – Y=g c) Y – X =g d) X = Y/g
3] A body is released from height h above the ground which takes ‘t’ seconds to reach the ground. The position of the body after t/2 seconds is
a) above the ground b) above the ground c) above the ground
d) Depends upon the size of the body
4] The average velocity of freely falling body is 21 ms-1. Then it is released from a height equal to
a) 5m b) 90m c) 30m d) 60m
5] A window is 0.5m high. A stone is released from a height 0.4m above the top of the window. The time taken by the stone to cross the window is
a) 4 s b) 3/7 s c) 1/7 s d) 2/7 s
6] A stone is dropped into a well and sound of the splash is heard 3 s later. If the depth of well is 44.1m find the velocity of sound in air (g = 9.8ms-2)
7] A body slides down a smooth inclined plane of inclination 300 with horizontal and length 19.6m starting from rest at the top. Find the speed and time to reach the bottom. (g = 9.8ms-2)
a) 2.5s, 15ms-1 b) 3.83s, 15.4ms-1 c) 2.83s, 13.87ms-1 d) None
8] A stone falls from the top of a tower in 8s. How much time will it take to cover the first quarter of distance starting from top?
a) 2s b) 3s c) 4s d) 8s
CLASS EXERCISE (B):
1] Two bodies one held 10m vertically above the other are released simultaneously. After falling freely for 3 seconds under gravity, their relative separation is
a) 10m b) 5m c) 1m d) none
2] After falling through the first h metres, a freely falling body acquires velocity ‘V’. After falling through the next h metres, velocity acquired by it would be
a) 2 V b) √2 c) V/√2 d) 4 V
3] A body is released from height ‘h’ above the ground. Exactly at the midway if g vanishes suddenly, its total time of fall is
a) b) 2 c) d)
4] A parachutist after bailing out (drops out) falls a distance of 19.6m after which the parachute opens and he decelerates downwards at 1.1ms-2. He reaches the ground with a speed of 2ms-1. Find total time taken and the height from the ground where parachutist bails out.
5] A ball is dropped from a height. Another ball is dropped from the same height after 2s. Their separation after 2 more second is
a) g b) 2g c) 4g d) 6g
6] A stone is dropped into water from a bridge 44.1m above the water. Another stone is thrown vertically downwards 1second later. Both strike the water simultaneously. Then initial speed of the second stone is
a) 24.5ms-1 b) 49 ms-1 c) 9.8 ms-1 d) 12.25ms-1
7] A stone is dropped from a multistored building. If it crosses 2 floors in the first second of its free fall, the numbers of floors it can cross in the 3rd second of its fall is
a) 5 b) 7 c) 10 d) 9
Body projected vertically upward CLASS EXERCISE (C):
1] A body is projected upwards with a velocity 98 m/s. Find: a) The maximum height reached
b) The time taken to reach the maximum height
c) Its velocity at a height 196 m from the point of projection d) Velocity with which it will cross down the point of projection
e) The time taken to reach back the point of projection
2] A boy throws a stone vertically up and catches it after time ‘t’ seconds. Then maximum height reached by the stone is
3] A body is projected vertically up with a velocity 19.6ms-1. Find the i) displacement after 4s ii) displacement in 2nd second. iii) Velocity after 3 seconds iv) ratio of displacements in 1st and 2nd second. iii) Velocity after 3 seconds iv) ratio of displacements in 1st and 2nd seconds respectively.
4] A body projected vertically up reaches a maximum height h in time t. The time taken to reach half of the maximum height is
a) t/2 b) t/4 c)
√ d) t 1 − √
5] A stone projected up with a velocity U reaches two points A and B at a distance ‘h’ with velocities U/2 and U/3. The maximum height reached by the stone is
a) b) c) d)
6] A body is projected vertically upwards. If t1 and t2 be the times at which it is at a height h above the point of projection while ascending and descending respectivley then
a) h = t1t2 b) h = 2gt1t2 c) h= d) h =
7] In above problem the velocity of projection is
a) g(t1+t2) b) g(t1+t2) c) 2g(t1 + t2) d) 4g(t1 + t2)
8] In above problem the maximum height reached by the body is a) g b) g c) g( + )2 d) g(t1+t2)2
9] In above problem the velocity of the body at height is
CLASS EXERCISE (D):
1] Assume that there is tower of sufficient height H. A particle is projected from bottom of tower with velocity u1 and another particle is dropped from top of tower with velocity u2. The time when the particles meet is T then
a) T =
( ) b) T = ( ) c) T =
( ) d) T = ( )
2] If a particle occupies x seconds less and acquires a velocity y ms----1 more at one place than at another in falling through the same distance. If g1 and g2 are accelerations due to gravity at those two places, then x: y is equal to
a) g1 / g2 b) g2 / g1 c) d) 1/
3] A rocket is fired vertically from the ground with a resultant vertical acceleration of 10 ms---2. The fuel is finished in one minute and the rocket continues to move up. If g = 10 ms---2, maximum height reached by the rocket is
a) 36 Km b) 18 Km c) 9 Km d) none
4] A body is projected vertically up from ground. In the last 2 seconds of its ascent distance travelled by it will be numerically
a) b) g c) 2g d) 4g
5] A body is projected vertically up from ground. In the last of its total journey distance travelled by it is h. Then its initial velocity of projection is numerically
6] Water drops fall from a tap on the floor 5m below at regular intervals of time. The first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be, from ground, at that instant when first strikes the ground, will be (g=10 ms-2).
a) 1.25 m b) 2.15 m c) 2.75 m d) 3.75 m Vertical projection from a height
CLASS EXERCISE (E):
1] A balloon is ascending up with a velocity of 4 ms--1. An object is dropped from it when it is at a height of 100 m above the ground. The distance between the object and balloon after 2 seconds is (g = 10 ms--2)
a) 10 m b) 20 m c) 30 m d) 40 m
2] A balloon rises from rest with a constant acceleration. A stone is released from it when it has risen to a height h. The time taken by the stone to reach the ground is
a) b) c)2 d) 4
3] From the top of a tower a stone is projected vertically up with a velocity 20 ms---1. After t seconds another3] from the top of a tower a stone is projected vertically up with a velocity 20 ms---1. After t seconds another stone is projected vertically down with a velocity 20 ms---1 so that both the stones reach the ground simultaneously. The t = (g = 10 ms--2)
4] A body projected vertically up with velocity u from the top of a tower reaches the foot of the tower with velocity 2u. Then height of that tower is
a) b) c) d)
5] A rocket fired vertically upwards with constant acceleration has its engine exhausted in 10 second. What is the maximum height reached by the rocket if its velocity at the end of 10th second 600 ms-1? [g = 10 ms---2]
a) 600 x 5 m b) 600 x 25 m c) 600 x 35 m d) 600 x 40 m
6] A particle is projected vertically upwards. Prove that it will be at of their greatest heights at times which are in the ratio 1: 3 7] A stone is released from a hydrogen balloon, going upwards
with velocity 12 m/s, when it is at height of 65m from the ground. Time the stone will take to reach the ground is
a) 5 sec b) 6 sec c) 7 sec d) 8 sec
8] From the top of a tower a stone is projected vertically upward. When it reaches a distance h below that point, its velocity is double that of its velocity when it was at a height ‘h’ above the top of the tower. Then greatest height attained by the stone above the top of the tower is
a) 2h b) h/3 c) 5h/3 d) 4h
9] A balloon rises from rest on the ground with a constant acceleration . A stone is dropped when the balloon has rises to a height of Hm. Find the time taken by the stone to reach the ground
SESSION – 13, 14 & 15 Aim:
Projectile -Motion in Two Dimensions
PROJECTILE -MOTION IN TWO DIMENSIONS
Motion in a plane is called as motion in two dimensions e.g., projectile motion, circular motion etc. For the analysis of such motion our reference will be made of an origin and two co-ordinate axes x and y. Position of particle is known by knowing its co-ordinate (x, y). Velocity of particle will be resultant of velocities in x and y direction Vx and Vy. Similarly acceleration will be in the two directions. For analysis of such motion we analyses the motion along two axes independently i.e., while dealing motion in x-direction we need not to think what is going on in y-x-direction and vice versa.
We have to study about projectile motion, circular motion and relative velocity under the head of motion in two dimensions.
Projectile Motion
A projectile is a particle, which is given an initial velocity, and then moves under the action of its weight alone. If the initial velocity is vertical, the particle moves in a straight line and such motion we had already discussed in ‘motion in one dimension as motion under gravity’. Here we are going to discuss the motion of particle which is projected obliquely near the earth’s surface. While discussing such motion we shall suppose the motion to be within such a moderate distance from the earth’s surface, that acceleration due to gravity may be considered to remain sensibly constant. We shall
also neglect the resistance of air and consider the motion to be in vacuum.
Important Terms used in projectile motion
When a particle is projected into air, the angle that the direction of projection makes with horizontal plane through the point of projection is called the angle of projection; the path, which the particle describes, is called the trajectory, the distance between the point of projection and the point where the path meets any plane draws through the point of projection is its range; the time that elapses in air is called as time of flight and the maximum distance above the plane during its motion is called as maximum height attained by the projectile.
Analytical treatment of projectile motion
Consider a particle projected with a velocity u of an angle with the horizontal from earth’s surface. If the earth did not attract a particle to itself, the particle would describe a straight line; on account of attraction of earth, however, the particle describes a curved path. This curve will be proved later to be always a parabola.
Let us take origin at the point of projection and x-axis and y-axis along the surface of earth and perpendicular to it respectively as shown in figure.
By the principle of physical independence of forces, the weight of the body only has effect on the motion in vertical direction. It,
y O P u x O
therefore, has no effect on the velocity of the body in the horizontal direction, and horizontal velocity therefore remains unaltered.
Motion in x- direction:
Motion in x - direction is motion with uniform velocity. At t = 0, x0 = 0 and ux = u cos
Position after time t, x = x0 + uxt ⟹ x = (u cos ) t
Velocity of any time t, vx = ux ⟹vx= u cos
Motion in y-direction:
Motion in y-direction is motion with uniformly acceleration When, t = 0, y0 = 0, uy = u sin and ay = -g
∴ After time‘t’, vy = uy + ay t ⟹vy = u sin - gt
y = y0 + uyt + ayt2 ⟹y = uyt + ayt2 y = (u sin ) t
-Also, = + 2ayy ⟹ = u2 sin2 - 2 gy
Time of Flight :
Time of flight is the time during which particle moves from O to O i.e., when t = T, y = 0
∴ From equation (iv)
O = u sin T- gT2 T = Range of projectile (R):
Range is horizontal distance travelled in time T. i.e., R = x (in time T)
∴ From equation (ii) R = ucos .T
= ucos R =
Maximum height reached (H):
At the time particle reaches its maximum height velocity of particle becomes parallel to horizontal direction i.e., vy = 0 when y = H
∴ From equation (v)
0 = u2 sin2 − 2 gH; H =
Equation of trajectory:
The path traced by a particle in motion is called trajectory and it can be known by knowing the relation between x and y.
From equation (1) and (2) eliminating time t we get y = x tan - sec2
This is trajectory of path and is equation of parabola. So we can say the path of particle is parabolic.
Velocity and direction of motion after a given time: After time‘t’ vx = ucos and vy = usin -gt
Hence resultant velocity v = +
= + ( − )
If direction of motion makes an angle with horizontal
Velocity and direction of motion at a given height:
At a height ‘h’, Vx = ucos
And Vy = − 2 ℎ
Resultant velocity v = + ; v = − 2 ℎ
Note that this is the velocity that a particle would have at height h if it is projected vertically from ground with u.
SOME IMPORTANT POINTS REGARDING PROJECTILE MOTION OVER A HORIZONTAL PLANE
i] For a given velocity of projection, the range of horizontal plane will be maximum when angle of projection is 45°.
We have range of projectile. R =
Therefore if we keep on increasing range will increase and then decrease. Its value will be maximum when sin2 is maximum i.e., = 450
Also, maximum range Rmax =
ii] For a given range and given initial speed of projection, there are two possible angle of projection which are complementary angle i.e., if one is other will be (900- ).Class Exercise solved Problem 1: The trajectory of a projectile is represented by y
=√3 − /2. The angle of projection is
a) 30o b) 45o c) 60o d) None of these
Sol: (c) By comparing the coefficient of x in given equation with standard equation
Problem 2: The path followed by a body projected along y-axis is given as by y = √3 – (1/2)x2 if g = 10 m/s, then the initial
velocity of projectile will be - (x and y are in m)
a) 3√10 m/s b) 2√10 m/s c) 10√3 m/s d) 10√2 m/s
Sol: (b) By comparing the coefficient of x2 in given equation with standard equation y = x tan − . = Substituting = 600 we get u =2√10m/s
Problem3: The equation of projectile is y = 16x– The horizontal range is
a) 16 m b) 8 m c) 3.2 m d) 12.8 m
Sol: (d) Standard equation of projectile motion y = x tan 1 − Given equation: y = 16x – or y = 16x 1 −
/ By comparing above equations R = =12.8 m.
Problem4: A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 Newton's in OF direction perpendicular to OE. The distance of the body from O after 4 seconds will be
a) 12 m b) 28 m c) 20 m d) 48 m
Sol: (c) Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds
∴ = = 3 × 4 = 12m and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds. ∴ = + = 4 =16m So its distance from O is given by d =
+ = (12) + (16) ∴ = 20
Problem 5: A body starts from the origin with an acceleration of 6 m/s2 along the x-axis and 8 m/s2 along the y-axis. Its distance from the origin after 4 seconds will be
a) 56 m b) 64 m c) 80 m d) 128 m
Sol: (c) Displacement along X- axis x = + = × 6 × (4) = 48
Displacement along Y- axis: y = + = × 8 × (4) = 48
Total distance from the origin = + = (48) + (64) = 80
Problem 6: In a projectile motion, velocity at maximum height is
a) b) c) d) None of these
Sol: (b) In a projectile motion at maximum height body possess only horizontal component of velocity i.e. u cos.
Problem7: A body is thrown at angle 30o to the horizontal with the velocity of 30 m/s. After 1 sec, its velocity will be (in m/s) ( = 10 / )
a) 10√7 b) 700√10 c) 100√7 d) √40
Sol: (a) From the formula of instantaneous velocity =
+ − 2
= (30) + (10) × 1 − 2 × 30 × 10 × 1 × 30 = 10√7 /
Problem 8: A projectile is fired at 30o to the horizontal. The vertical component of its velocity is 80 ms-1. Its time of flight is T. What will be the velocity of the projectile at t = T/2
a) 80 ms-1 b) 80/√3ms-1
Sol: (b) At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only.
Given uvertical =usin = 80 ⟹ = = 160 / ∴ horizontal= ucos = 160cos 30 = 80√3 /
Problem 9. A particle is projected from point O with velocity u in a direction making an angle with the horizontal. At any instant its position is at point P at right angles to the initial direction of projection. Its velocity at point P is
a) u tan b) u cot c) u cosec d) u sec Sol: (b) Horizontal velocity at point ‘O’ = ucos
Horizontal velocity at point ‘P’ = v sin
In projectile motion horizontal component of velocity remains constant throughout the motion
∴ v sin = ⟹ = 90o u v O P 90o u v O P v sin 90o – u s in u cos
Problem 10: A particle P is projected with velocity u1 at an angle of 30o with the horizontal. Another particle Q is thrown vertically upwards with velocity u2 from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is
a) = b) = 2 c) = d) = 4
Sol: (b) Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal
30 = ⟹ = ∴ = 2
Problem 11.Two seconds after projection a projectile is travelling in a direction inclined at 30o to the horizontal after one more sec, it is travelling horizontally, and the magnitude and direction of its velocity are
a) 2√20 / , 60 b) 20√3 / , 60 c) 6√40 / , 30 d) 40√6 / , 30
Sol: (b) Let in 2 sec body reaches upto point A and after one more sec upto point B. Total time of ascent
u2 u1 30o P Q u O B v A 30o u cos
For a body is given 3 sec i.e. t = =3 ∴ = 10 × 3 = 30 ….. (i)
Horizontal component of velocity remains always constant ucos = 30 ... (ii)
For vertical upward motion between point O and A
30 = − × 2[ = − ]
30 = 30 − 20[ = 30] ∴ = 20 / Substituting this value in equation (ii)
ucos = 20 30 = 10√3….. (iii)
From equation (i) and (iii) and u =20√3 = 60
Problem12: A body is projected up a smooth inclined plane (length =20√2m) with velocity u from the point M as shown in the figure. The angle of inclination is 45o and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of .
a) 40ms–1 b) 40√2 ms–1 c) 20 ms–1 d) 20√2 ms–1
Sol: (d) At point N angle of projection of the body will be 45°. Let velocity of projection at this point is v. If the body just manages to cross the well then
40 m
45o
= 40 [as = 45 ] = 400⟹ v = 20 m/s
But we have to calculate the velocity (u) of the body at point M. For motion along the inclined plane (from M to N) Final velocity (v) = 20 m/s, acceleration (a) = - g sin = - g sin 45o, distance of inclined plane =
20√2(20) − 2
√ . 20√2[Using v2 = u2 + 2as] = 20 + 400 ⟹ = √ .
Problem 13: A projectile is fired with velocity u making angle with the horizontal. What is the change in velocity when it is at the highest point?
a) ucos b) u c) u sin d) (u cos - u)
Sol: (c) Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes. Initially vertical component usin .Finally it becomes zero. So change in velocity = u sin
Problem14: A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30o with the horizontal. The change in momentum (in magnitude) of the body is
a) 24.5 N-s b) 49.0 N-s c) 98.0 N-s d) 50.0 N-s 40 m 45o M R N u v
Sol: (b) Change in momentum between complete projectile motion = 2musin =2×0.5×98×sin300=49 N-s.
Problem15: A particle of mass 100 g is fired with a velocity 20 m sec-1 making an angle of 30o with the horizontal. When it rises to the highest point of its path then the change in its momentum is
a) √3 kg m sec–1 b) 1/2 kg m sec-1
c) √2 kg m sec–1 d)1 kg m sec-1
Sol: (d) Horizontal momentum remains always constant
So change in vertical momentum ( Δ ⃗ ) = Final vertical momentum - Initial vertical momentum
= 0– mu sin
|ΔP| = 0.1 × 20 × 30 = 1 /
Problem 16: Two equal masses (m) are projected at the same angle ( ) from two points separated by their range with equal velocities (v). The momentum at the point of their collision is a) Zero b) 2 mv cos c) - 2 mv cos d) None
Sol: (a) Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero.
v v
Problem17: A particle of mass m is projected with velocity making an angle of 45o with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)
a) Zero b) mv3/ (4√2 )
c) mv3/(√2 ) d) mv2/2g
Sol: (b) = L = =
√ [As = 45o]
Problem18: A body is thrown with a velocity of 9.8 m/s making an angle of 30o with the horizontal. It will hit the ground after a
time
a) 1.5 s b) 1 s c) 3 s d) 2 s Sol: (b) T = = × . ×
. = 1sec
Problem 19: Two particles are separated at a horizontal distance as shown in figure. They are projected at the same time as shown in figure with different initial speed. The time after which the horizontal distance between the particles become zero is
a) u/2x b) x/ u c) 2u / x d) u /x
Sol: (b) Let x1 and x2 are the horizontal distances travelled by particle A and B respectively in time t.
x1 = √ . 30 × ... (i) and x2= 60 × ... (ii) x1+x2= √ . 30 × + 60 × = ∴ = / Problem 20: A particle is projected from a point O with a velocity
in a direction making an angle upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is
Sol: (b) When body projected with initial velocity by making angle with the horizontal. Then after time t, (at point P) its direction is perpendicular to.
Magnitude of velocity at point P is given by = (from sample problem no. 9).For vertical motion: Initial velocity (at point O) = usin
Final velocity (at point P) = –v cos = − Time of flight (from point O to P) = t
Applying first equation of motion = − − = usin −
= =
[ + cos ] =
Problem 21: A ball is projected upwards from the top of tower with a velocity 50 ms-1 making angle 30o with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground?
a) 2.33 sec b) 5.33 sec c) 6.33 sec d) 9.33 sec
Sol: (c) Formula for calculation of time to reach the body on the ground from the tower of height 'h' (If it is thrown vertically up with velocity u) is given by t = 1 + 1 + So we can resolve the given velocity in vertical direction and can apply the above formula. Initial vertical component of velocity
90o u P O v (90 – ) v cos u s in u cos
U sin = 50 sin30 = 25 m/s. t =
. 1 + 1 +
× . ×
( ) =6.33 sec
Problem 22: If for a given angle of projection, the horizontal range is doubled, the time of flight becomes
a) 4 times b) 2 times c) √2 times d) 1/√2times Sol: (c) = and T =
∴ and T (If and g are constant).
In the given condition to make range double, velocity must be increased upto √2 times that of previous value. So automatically time of flight will becomes √2 times.
Problem 23. A particle is thrown with velocity u at an angle from the horizontal. Another particle is thrown with the same velocity at an angle from the vertical. The ratio of times of flight of two particles will be
a) tan 2 : 1 b) cot 2 : 1 c) tan : 1 d) cot : 1
Sol: (c) For first particles angle of projection from the horizontal is So T1=
For second particle angle of projection from the vertical is . It mean from the horizontal is (90- )
∴ T1=
( )
=
Problem 24: The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = 10 ms-2). The time of flight will be decreased by
a) 0% b) 1% c) 9% d) 11%
Sol: (c) T = ∴ = = =
Fractional decrease in time of flight = = Percentage decrease = 9%
Problem25: A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30o with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground (g = 10 m/s2, sin 30o =
sin 30o = √ )
a) 8.66 m b) 5.20 m c) 4.33 m d) 2.60 m Sol: (a) Simply we have to calculate the range of projectile
R = = ( ) ( × ) R=5 √3
30o
10 m
u
SESSION – 16 & 17 Aim:
Projection on Inclined Plane. Horizontal projection
PROJECTION ON INCLINED PLANE.
A particle is projected from a point A on an inclined plane, which is inclined at an angle to the horizon with a velocity at an elevation. The direction of projection lines in the vertical plane through AB, the line of the greatest slope of the plane.
Let the particle strike the plane at B so that AB is the range on the inclined plane.
The initial velocity of projection u can be resolved into a component u cos ( - ) along the plane and a component u sin ( - ) perpendicular to the plane. The acceleration due to gravity g which acts vertically down can be resolved into components g sin up the plane and g cos perpendicular to the plane. By the principle of physical independence of forces the motion along the plane may be considered independent of the motion perpendicular to the plane. Let T be the time, which the particle takes to go from A to B. Then in this time the distance traversed by the projectile perpendicular to the plane is zero.
∴ 0 = u sin ( - ) T - g cos T2 T = ( )
u
A
B C During this time the horizontal velocity of the projectile (u cos ) remains constant. Hence the horizontal distance described is given
by AC = u cos T = ( ) ∴ = = ( )
Range on the inclined plane = ( )
Maximum range on the inclined plane
R = ( )
= [sin (2 - ) - sin ]
For given values of u and , R is maximum when sin (2 − ) = 1 i.e., (2 − ) = 90°
= (450 + /2)
If Rm represents the maximum range on the inclined plane. Rm = (1- sin ); Rm = ( )
For a given velocity of projection, it can be shown that there are two directions of projection which are equally inclined to the direction of maximum range.
Now R =
[sin (2 - ) - sin ]
For given values of u, and R, sin (2 − ) is constant. There are two values of (2 − ) each less than 1800 that can satisfy the above equation.
Let (2+
_ − ) and (2 − ) be the two values. Then 2 - = 1800 - (2 - ) ; - /2 = 900 - ( - /2)
- (450 + /2) = (450 + /2) -
Since (45° + /2) is the angle of projection giving the maximum range, it follows that the direction giving maximum range bisects
