Example 1:
Water drops are falling at regular intervals from a roof. At an instant when a drop is about to leave the roof find the ratio of separations between 3 successive drops below the roof.
Solution :
Let the drops fall at a regular interval of ‘t0’ from the roof.
When a drop is about to leave the roof the 1st, 2nd and 3rd drop below it have travelled for time interval t0, 2t0 and 3t0 respectively.
Distance below the roof 1st drop → 1
A body is in straight line motion with an acceleration given by a = 32 – 4v. Find the velocity of the body at t = ln 2 sec if initially its velocity was 4 m/sec.
Solution : On integrating with appropriate limits we have
432 4
S The acceleration of a point started at t = 0, varies with time as shown in 5.0 given graph. Find the distance travelled in 30 seconds and draw the velocity-time and position-time graphs if initial velocity and position are zero. As given in the acceleration-time graph,
for time interval (0, 10) : a is constant and +ve
⇒ velocity increases linearly.
⇒ slope of x – t graph must be increasing for time interval (10, 20) : a is zero
⇒ velocity is constant
⇒ slope of x – t graph is constant for time interval (20, 30) : a is constant and –ve
⇒ velocity decreases linearly.
⇒ slope of x – t graph is decreasing At t = 10 s, v = u + at = 0 + 5 × 10 = 50 m/s.
As we see from v – t graph, velocity is always nonnegative for time interval (0, 30), therefore distance travelled is equal to displacement.
s = displacement = area under v – t graph
An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.
(g = 32 ft/s2) Initial velocities of both elevator and coin are zero. Suppose that the coin
falls on floor in t seconds, then the distance travelled by coin = distance travelled by floor + initial distance between coin and floor
1
2gt2 = 6 + 1 2at2 Substituting t = 1
we get g = 12 + a
⇒ a = 32 – 12 = 20 ft/s2.
Example 5: Two ships, 1 and 2, move with constant velocities 3 m/s and 4 m/s along two
mutually perpendicular straight tracks towards the intersection point O. At the moment t = 0 the ships were located at the distances 120 m and 200 from the point). How soon will the distance between the ships become the shortest and what is it equal to?
Solution :
Suppose that ship 1 starts moving from point B (at t = 0) and ship 2 from point A (at t = 0 only), then after time ‘t’, 1 must have travelled 4t units along –ve x–axis. Therefore, at some ‘t’ ship 1 is at point (– (200 – 4t), 0) and ship 2 is at point (0, 120 – 3t).
Therefore, distance between 1 and 2 is d =
(
x1−x2) (
2+ y1−y2)
2Therefore, derivative of d2 w.r.t. time must be zero.
∴ 2(200 – 4t)(– 4) + 2(120 – 3t)(– 3) = 0
⇒ – 1600 + 32t – 720 + 18t = 0
⇒ 50t = 2320 ⇒ t = 2320/50 = 46.4 sec.
Substituting t = 46.4 sec in equation (i), we will get the minimum distance, which comes out to be 24 m.
Example 6:
A rubber ball is released from a height of 490 m above the floor. It bounces repeatedly, always rising to 81/100 of the height through which it falls.
(a) Ignoring the practical fact that the ball has a finite size (in other words, treating the ball as point mass that bounces an infinite number of times), show that its total distance of travel is 46.7 m.
(b) Determine the time required for the infinite number of bounces.
(c) Determine the average speed. (Take g = 9.8 m/s2)
(b) Time required to fall through height h = 2h g
(c) Average speed = distance travelled A particle is projected with speed u at an angle ‘α’ with horizontal.
Find the speed of the particle at the moment when the velocity makes an angle ‘β’ with horizontal.
Solution :
Let the speed of the particle at the instant when the velocity vector makes an angle ‘β’ be v.
Horizontal component of velocity at that instant = v cosβ Horizontal component of intial velocity = u cosα.
Equating the horizontal components we get u cosα = v cosβ [Q ax = 0]
or v = ucosβcos α
Example 8:
A ship is approaching a cliff of height 105 m above sea level. A gun fitted on the ship can fire shots with a speed of 110m/sec. Find the maximam distance from the foot of the cliff from where the gun can hit an object on the top of the cliff. [g = 10m/s2]
Solution :
Let that distance be ‘D’, θ be the angle of projection and ‘T’ be the time taken to hit the cliff.
For Horizontal Motion :
(110 cosθ)T = D ———— (1)
For Vertical Motion :
105 m
Substituting ‘T’ From eqution (1) we have 105 =
Differentiating the above equation w.r.t ‘θ’ we get,
0 = 2 2 2 2 2
Putting D tanθ = 1210 in equation (3) we have
105 = 1210 – (110)5 2×
(
D2+(1210 )2)
⇒ D = 1100 mExample 9:
A vertical pole has a red mark at some height. A stone is projected from a fixed point on the ground. When projected at an angle of 45° it hits the pole perpendicularly 1 m above the mark. When projected with a different velocity at an angle of 37°, it hits the pole perpendicularly 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark perpendicularly. [g = 10 m/s2]
Solution :
The stone hits the pole perpendicularly means that the stone was at highest point of trajectory.
When angle of projection = 45°
2D = Range and H + 1 = Max. Height
When angle of projection = 37°.
37°45° Solving these equation we get u = 3620
3 m/sec and θ = tan–1 9 Find the range ‘R’ of a particle projected with speed u perpendicular
to the plane of inclination θ with horizontal.
Solution :
Here ux = 0, uy = u
ax = g sinθ, ay = –g cosθ Let Range be ‘R’ and time of flight be ‘T’.
y-coordinate will become zero when projectile will land on the plane.
applying sy = uyt + 1
Example 11:
A ferrari, mercedes and a toy rocket are moving as shown in the figure. Assuming x-y plane to be the plane of motion of objects.
Find
(a) Velocity of Ferrari w.r.t Rocket.
FERRARI MERCEDES
15 m/sec 20 m/sec
ROCKET
37°
30 m/sec y
x
(b) Velocity of Mercedes w.r.t Ferrari.
(c) Velocity of tree w.r.t rocket.
(d) Velocity of mercedes w.r.t rocket.
Solution :
A standing man, observes rain falling with velocity of 20 m/sec at an angle of 30° with the vertical.
(a) Find the velocity with which the man should move so that rain appears to fall vertically to him.
(b) Now if he further increases his speed, rain again appears to fall at 30° with the vertical. Find his new velocity.
Solution :
VR, G
ur
= −20sin 30° −iˆ 20 cos 30°ˆj
= −10iˆ−10 3ˆj.
(a) Let the man is moving with speed V0 towards negative x-axis.
(In this case only he will feel the rain coming down vertically)
∴ VM , G
ur
= −Vur0iˆ
∴ VR, M
ur
= Velocity of rain w.r.t man
= VR , G
Rain feels to come down vertically to man.
∴ – 10 + V0 = 0
⇒ V0 = 10 m/sec
∴ VurM , G = −10iˆ m/sec
(b) Let the new increased speed of man is V'.
S
∴ VurM, G = V i′ˆ
makes an angle = 30° with vertical.
⇒ tan 30° = V 10 10 3
′ − ⇒ V' = 20 m/sec
Example 13:
There are particles A, B and C are situated at the vertices of an equilateral triangle ABC of side a at t = 0. Each of the particles moves with constant speed V. A always has its velocity along AB, B along BC and C along CA. At what time will the particle meet each other?
Solution :
The motion of the particles is roughly sketched in figure. By symmetry they will meet at the centroid O of the triangle. At any instant the particles will from an equilateral triangle ABC with the same Centroid O. All the particles will meet at the centre. Concentrate on the motion of any one particle, say B. At any instant its velocity makes angle 30° with BO. The component of this velocity along BO is V cos 30°. This
component is the rate of decrease of the distance BO.
O
a = displacement of each particle. Therefore, the time taken for BO to become zero
C
A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s2 and the projection speed in the vertical direction is 9.8 m/sec. How far behind the boy will the ball fall on the car?
Solution :
Distance travelled by car in Horizontal direction = xc
= u × 2 + 1
2 × 1 × 22 = 2u + 2
Distance travelled by ball in Horizontal direction = xBall = 2 × u = 2u
∴xc – xB = 2m
Example 15:
An elevator is going up with an upward acceleration of 1m/s2. At the instant when its velocity is 2m/sec, a stone is projected upward from its floor with a speed of 2m/sec relative to the elevator, at an elevation of 30°.
(a) Calculate the time taken by the stone to return to floor.
(b) If the elevator was moving with a downward acceleration equal to g, how would the motion be.
Solution :
(a) Initial velocity of Ball w.r.t lift = 2 cos 30° +iˆ 2sin 30°ˆj
= 3iˆ+ˆj Initial velocity of Lift = ˆ2 j
Initial velocity of ball w.r.t ground = VurB G, =VurB L, +VurL G,
= 3iˆ+3ˆj
The ball will return to floor when y displacement of floor of lift becomes equal to y displacement of ball.
yBall = uyT + ayT2 = 3T – 1
2 × 10 × T2 ylift = 2T
Equating both we get 3T – 5T2 = 2T ⇒ T = 0.2 sec
If lift is moving with downward acceleration ‘g’ then relative acceleration of stone w.r.t lift = aS,G −aL,G
= − − −gjˆ ( gjˆ) = 0 Also VB,L = 3iˆ+3ˆj
∴ Ball will keep on separating from floor of lift with velocity VB,L= 3iˆ+3ˆj Motion of ball w.r.t someone sitting in lift will be straight line.
S O L V E D
E X A M P L E S