Law of Conservation of Mometum
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Calculate the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25 g acquires the velocity of 500 ms−1 after firing from the gun.
- 10 ms−1
25 ms−1
- 2.5 ms−1
- 5 ms−1
- 24 ms−1 towards west
- 14 ms−1 towards east
- 24 ms−1 towards east
- 54 ms−1 towards south
- 10 ms−1
- 20 ms−1
- 30 ms−1
- 50 ms−1
- 1.81 ms−1
- 1.3 ms−1
- 1.92 ms−1
- 2.44 ms−1
- 0.75 ms−1
- 0.75 kms−1
- 0.75 cms−1
- 0.57 ms−1
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms−1 with respect to the ground. Calculate the initial recoil velocity of the rifle.
0.4375 m/s
0.4 m/s
0 m/s
0.3375 m/s
Ball A of mass 5 kg travelling at 6 ms−1collides with another ball B of mass 2 kg travelling at 2 ms−1 in the same direction. After collision, the velocity of ball A reduces to 2 ms−1. Calculate the velocity of ball B after the collision.
10 ms−1
12 ms−1
6 ms−1
8 ms−1
The total momentum of a system always remains unchanged before and after the collision.
- True
- False
- v
- 2v
- v/2
- ∞
- 10 kgms−1
- 1.5 kgms−1
- 2.25 kgms−1
- 2.5 kgms−1
A bullet of mass 20g moving with a velocity of 100 ms−1 strikes a wooden block of mass 80g at rest and gets embedded into it. Calculate the velocity of the combined system.
10 ms-1
30 ms-1
20 ms-1
5 ms-1
Explain how the law of conservtion of momentum can be used to explain
(i) recoil of a gun
(ii) motion of a rocket. [3 MARKS]
A bullet was fired from a gun. Choose the correct option for the situation assuming that no external force is acting on the gun-bullet system.
Since the mass of bullet is very less than the mass of the gun, the momentum of bullet is not equal to the momentum of gun.
The magnitude of momentum of gun is equal to that of bullet.
During firing, the law of conservation of momentum does not hold true.
The momentum of gun is twice the momentum of bullet.
0.10 ms−1
- 0.35 ms−1
- 0.15 ms−1
- 0.25 ms−1
- (√MM+m)v
- (√mM+m)v
- (mM+m)v
- (m+MMm)v
- 5 ms−1
- 10 ms−1
- 20 ms−1
- 4 ms−1
Ball A of mass 5 kg travelling at 6 ms−1collides with another ball B of mass 2 kg travelling at 2 ms−1 in the same direction. After collision, the velocity of ball A reduces to 2 ms−1. Calculate the velocity of ball B after the collision.
10 ms−1
12 ms−1
6 ms−1
8 ms−1
- m1v2+m2v1
- m1u2+m2u1
- m1u1+m2u2
- m1u1+m2v2
- 5.5ms−1
- 2.25ms−1
- 2.75ms−1
- 7.5ms−1
- one direction and with equal velocities.
- opposite directions with equal velocities.
- opposite directions with different velocities.
- one direction with different velocities.