Collisions(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

COLLISIONS MCQs

Total Questions : 29 | Page 3 of 3 pages

Question 21. A body of mass m₁ moving at a constant speed undergoes an elastic collision with a body of mass

m_{2}

initially at rest. The ratio of the kinetic energy of mass

m_{1}

after the collision to that before the collision is

(m1−m2m1+m2)2
(m1+m2m1−m2)2
(2m1m1+m2)2
(2m2m1+m2)2

Discuss Question

Answer: Option A. -> (m1−m2m1+m2)2
:
A
Let

u_{1}

be the speed of mass

m_{1}

before the collision.
Here

u_{2} = 0

. Therefore, the speeds of masses

m_{1}

and

m_{2}

after the collision respectively are

v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1}

and

v_{2} = (\frac{2 m_{1}}{m_{1} + m_{2}}) u_{1}

∴

KE of

m_{1}

after collision =

\frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} u_{1}^{2}

. KE of

m_{1}

before collision =

\frac{1}{2} m_{1} u_{1}^{2}

.
The ratio of the two is

{(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2}

. Hence the correct choice is (a)

Question 22. A completely inelastic collision is one in which the two colliding particles

Are separated after collision
Remain together after collision
Split into small fragments flying in all directions
None of the above

Discuss Question

Answer: Option B. -> Remain together after collision
:
B
They continue to remain together as the time taken to complete is nearly infinity

Question 23. A spacecraft of mass 'M' explodes into two parts when its velocity is 'V'. After the explosion, one part of mass 'm' is left stationary. What is the velocity of the other part?

MV(M+m)
MV(M−m)
mV(M−m)
MV(m)

Discuss Question

Answer: Option B. -> MV(M−m)
:
B
Mass of the second part = M – m. If its velocity is v, then we have MV = (M – m) v + m

\times

0
or

v = \frac{M V}{(M - m)}

, which is choice (b).

Question 24. A body of mass 'm' moving with a constant velocity 'v' hits another body of the same mass moving with the same velocity 'v' but in the opposite direction and sticks to it. The velocity of the compound body after collision is

v
2v
Zero
v2

Discuss Question

Answer: Option C. -> Zero
:
C
Initial momentum of the system = mv - mv = 0
As body sticks together ;final momentum = 2mV
By conservation of momentum 2mV = 0.
V = 0

Question 25. A ball of mass 'm' moving horizontally at a speed 'v' collides with the bob of a simple pendulum at rest. The mass of the bob is also 'm'. If the collision is perfectly inelastic, the ratio of the kinetic energy of the system immediately after the collision to that before the collision will be

1 : 1
1 : 2
1 : 3
1 : 4

Discuss Question

Answer: Option B. -> 1 : 2
:
B
Mass of the ball and the bob sticking together is

m^{^{'}} = 2 m

.
KE after collision

= \frac{1}{2} m^{^{'}} v^{^{'} 2} = \frac{1}{2} \times 2 m \times {(\frac{v}{2})}^{2} = \frac{1}{4} m v^{2}

.
KE before collision =

\frac{1}{2} m v^{2}

.Therefore, their ratio is 1:2. Hence the correct choice is (b).

Question 26. A body of mass 2 kg moving with a velocity

(^i + 2^j - 3^k) m s^{- 1}

collides with another body of mass 3 kg moving with a velocity

(2^i +^j +^k)

m s^{- 1}

. If they stick together, the velocity in

m s^{- 1}

of the composite body is

15(8^i+7^j−3^k)
15(−4^i+^j−3^k)
15(8^i+^j−^k)
15(−4^i+7^j−3^k)

Discuss Question

Answer: Option A. -> 15(8^i+7^j−3^k)
:
A

p_{1} = 2 k g (^i + 2^j - 3^k) m s^{- 1} = (2^i + 4^j - 6^k) k g m s^{- 1} p_{2} = 3 k g (2^i +^j +^k) m s^{- 1} = (6^i + 3^j + 3^k) k g m s^{- 1}

Resultant momentum is

p = p_{1} + p_{2}

(2^i + 4^j - 6^k) + (6^i + 3^j + 3^k)

(8^i + 7^j - 3^k) k g m s^{- 1}

Total mass (m) = 2 + 3 = 5 kg. Therefore, the velocity of composite body is

v = \frac{p}{m} = \frac{1}{5} (8^i + 7^j - 3^k) m s^{- 1}

Hence the correct choice is (a).

Question 27. A neutron of mass

1.67 \times 10^{- 27} k g

is moving with a velocity

1.2 \times 10^{7} m s^{- 1}

collides head-on with a deuteron of mass

3.34 \times 10^{- 27} k g

initially at rest. If the collision is perfectly inelastic, the speed of the composite particle will be

2×106ms−1
4×106ms−1
6×106ms−1
8×106ms−1

Discuss Question

Answer: Option B. -> 4×106ms−1
:
B
Mass of neutron (m) =

1.67 \times 10^{- 27}

kg. Speed of neutron (v) =

1.2 \times 10^{7} m s^{- 1}

.Notice that the mass ofdeuteron (M) =

3.34 \times 10^{- 27} k g = 2 m

. If V is the speed of the composite particle, the law of conservaton of momentum gives mv = (m+M)V
or

V = \frac{m v}{m + M} = \frac{m v}{m + 2 m}

V = \frac{v}{3} = \frac{1.2 \times 10^{7}}{3} = 4 \times 10^{6} m s^{- 1}

, which is choice (b).

Question 28. A lead ball strikes a wall and falls down, a tennis ball having the same mass and velocity strikes the wall and bounces back. Check the correct statement

The momentum of the lead ball is greater than that of the tennis ball
The lead ball suffers a greater change in momentum compared with the tennis ball
The tennis ball suffers a greater change in momentum as compared with the lead ball
Both suffer an equal change in momentum

Discuss Question

Answer: Option C. -> The tennis ball suffers a greater change in momentum as compared with the lead ball
:
C
Change in the momentum
= Final momentum - initial momentum

A Lead Ball Strikes A Wall And Falls Down, A Tennis Ball Hav...

For lead ball

△ {\to P}_{l e a d} = 0 - m \to v = - m \to v

For tennis ball

△ {\to P}_{t e n n i s} = - m \to v - m \to v = - 2 m \to v

i.e. tennis ball suffers a greater change in momentum.

Question 29. A heavy steel ball of mass greater than 1 kg moving with a speed of 2

m s e c^{- 1}

collides head on with a stationary ping-pong ball of mass less than 0.1 gm. The collision is elastic. After the collision the ping-pong ball moves approximately with speed

2m sec−1
4m sec−1
2×104m sec−1
2×103m sec−1

Discuss Question

Answer: Option B. -> 4m sec−1
:
B
We know that when heavier body strikes elastically with a lighter body then after collision lighter body will move with double velocity that of heavier body.i.e.the ping pong ball move with speed of