Centre of Mass, Conservation of Linear Momentum, Impulse
15. Assertion is false as explained in the answer to question no. 12
2
2 as given is Reason which is true.
Option (d) is correct.
11. As no external force would be acting on the system of electron and proton along the line joining electron and proton the CM of electron and proton will remain at rest. Therefore, Assertion is false.
Further, as proton is heavier than electron the reason is true.
Option (d) is correct.
12. v m m
m m v m
m m v
1 1 2
1 2
1 2
1 2
2
′ = − 2
+ +
+
and v m m
m m v m
m m v
2 2 1
1 2
2 1
1 2
1
′ = − 2
+ +
+ v2′ −v1′
= − −
+ + − +
+
m m m
m m v m m m
m m v
2 1 2
1 2
2 1 1 2
1 2
1
2 2
= −v2+v1 = −(v2−v1) i.e., v′21= −v21
i.e., relative velocity of A w.r.t. B after collision
= − (relative velocity of A w.r.t. B before collision)
∴ Reason is true and Assertion is false.
Option (d) is correct.
13. As explained in the answer to question no.
11, the CM of the objects will remain at rest. Therefore, assertion is false.
m x1 1=m x2 2
∴ x
x m m
2 1
1 2
=
∴ x2>x1 as m2>m1 Reason is true.
Option (d) is correct.
14. F→ dp→
= dt (Newton's second law of motion)
a F
→ →
= m (outcome of the above)
∴ Reason is true.
First equation tells that if same force F→ is applied on different masses the rate of change of momentum i.e., d→p
dt of each mass will be same. Second equation tells that if same force F→ is applied on different masses the a→ produced in each will be different.
∴ Assertion is true.
Further, as Reason is the correct explanation of the Assertion.
Option (a) is correct.
15. Assertion is false as explained in the answer to question no. 12.
In every type of collision the linear momentum of the system remains conserved. Therefore, Reason is true.
Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 189
v2
Before collision
v '2
m2 m1
m1 m2
v1
A B
After collision v '1
B A
x1 x2
m1 CM m2
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m v Option (b) is correct.
2. mu= 2mv Option (c) is correct.
3. Here |→v2| | |= v→1 = v say
Net momentum of the two elements as shown in figure =dp |= dp→| Option (b) is correct.
4. 1
Option (b) is correct.
l
5. As the collisions of the striker and the walls of the carrom are perfectly elastic the striker will follow the path OPQROP ...
Change in KE = work done against friction 1
2mv2−0= µmgs
(m = mass of striker, s = displacement of the striker)
∴ Striker will stop at point Q where co-ordinates are 1
2 2
Option (a) is correct.
6. As no force is acting on the system along horizontal, the CM of the system will not shift horizontally.
4p=1 1( − p)
⇒ p =1
5 m
Displacement ( )x of bar when pendulum becomes vertical
x When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m.
Option (b) is correct.
7.
Momentum imparted to the floor in 1st collision = p− −( ep)= p(1+e)
2nd collision =ep− −( e p2 )=ep(1+e) 3rd collision =e p2 − −( e p3 )=e p2 (1+e) As theoretically there will be infinite collision, total momentum imparted to floor Option (d) is correct.
8. Let F be the frictional force applied by plate when bullet enters into it
∴ 1
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 191
x
= − + Option (a) is correct.
9. Option (b) is correct.
10. y
12. As the resultant of the velocities of 1st and 2nd are just opposite to that of 3rd, the 4th particle will travel in the line in which 3rd is travelling.
Let the velocity of 4th particle is u as shown in figure.
∴ ucos45° +vcos45° =v i.e., u=v( 2−1) Total energy released
=1 + + + −
C will increase the tension in the string.
mu= 2mv2 i.e., v u
2=2
C will also increase the weight of B when collision takes place. Thus, mu=(2m+m v) 3
∴ v v1: 2:v3
=u u u
2 2 3: : = 3 3 2: : Option (b) is correct.
14.
vCM=vcos 30° = v 3
2 Option (a) is correct.
15. →rCM=i^ r1
→ (position vector of lighter piece)
=3^i+2^j−4k^
r r r
→ → →
= +
CM +
m m
m m
1 1 2 2
1 2
r r r
2 1 2 1 1
2
→ → →
=(m +m ) −m m
CM
=
− + −
2 2
3 3 2 4
4 3
i i j k
^ ^ ^ ^
( )
=1 − + −
4[6^i 2 3( ^i 2^j 4k^)]
=1 − + 4[ 4^j 8k^] = − +^j 2k^
∴ The heavier part will be at ( ,0 −1 2, ). Option (d) is correct.
16. Motion of A : h gt 2
1 2
= 2, vA (at time t) = g t= gh
i.e., t h
= g
Motion of B :
h v h
g gh
g 2
1
= −2
i.e., h v h
= g
∴ v= gh
v gh g h
B= − ⋅ g = 0
Collision of A and B at time t :
m gh= 3mω
∴ ω = gh
3
Velocity of the combined mass when it reach ground
v′ =2 ω2+2gh = gh+
g 2gh
i.e., v gh
′ = 19 3 Option (d) is correct.
17. u = velocity of man w.r.t. cart Let v = velocity of cart w.r.t. ground
∴ Velocity of man w.r.t. ground =u+v m u( +v)+2mv=3m⋅0
∴ v u
= −3
Work done = KE gained by man and cart
=1 + +
2
1 22
2 2
m u( v) mv
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 193
v 30°
v cos 30°
CM h/2 h/2
h
Place of collision
w 3 m m A
time = t (say)
v 2m
B
g
A B A B
m + 2m m 2m
gh Rest ω
= −
Option (d) is correct.
18. v m m
If the velocity of CM becomes zero at displacements
02=402+2(−10) s
⇒ s = 80 m
∴ Maximum height attained by CM
=20m+80m = 100 m Option (c) is correct.
19. As the masses are equal and the collision is elastic, the particles will exchange their velocities as shown in figure.
Gain in KE of 1st particle =1 − − Option (c) is correct.
20. According to question
(4 ) (4 ) ( ) ( ) Option (b) is correct.
21. xCM of Fig. 1 will as that of Fig. 2.
Initial position of CM 30 m/s
v 2v
B A
In terms of velocity
2v v
In terms of momentum
Before collision
y
There is no need to find the value of yCM
Option (a) is correct.
22. Conservation of momentum along y-axis m v Conservation of momentum along x-axis.
m v
Squaring and adding Eqs. (i) and (ii), 2v v20 2
=
∴ v v
= 0 2 2 Option (b) is correct.
23. 1
(When wall just breaks off the velocity of mass m1 would be zero)
Option (b) is correct.
24. Velocity of 2nd ball when 1st with velocity v strikes 2nd at rest
Velocity of 3rd ball when 2nd with velocity 4
3v strikes 3rd at rest
v m As in every collision m
m m for it to complete the circle
∴ 4
Option (a) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 195
m0
25. Impulse given to the block will also release it from abstraction besides giving and then imparting the restoring force on it due to velocity to it 5cm of the spring expansion will accelerate it.
Impulse =4kg ms−1
Initial velocity ( )u =4 kg ms = 2 kg
–1
2 ms−1 Average Acceleration ( )a
= =
×
× kx
m 2
4000 5
100 2 2 = 50 m/s2 Displacement ( )s =x= 5
100 m
∴ v2=u2 +2as =2 +2×50× 5
100
2
=4+5
= 9
∴ v =3ms−1
Option (b) is correct.
26. Compression in spring Velocity gained