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Question: Answered & Verified by Expert
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m s-1 in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is

PhysicsLaws of MotionNEET
Options:
  • A 250 m s-1
     
  • B 350 m s-1
  • C 400 m s-1
  • D 500 m s-1
Solution:
2085 Upvotes Verified Answer
The correct answer is: 500 m s-1
Time of flight tf=2hg=2×510=1 s
Horizontal range (R) = horizontal velocity × time of flight
Horizontal velocities of the bullet and of the ball after the collision respectively are
vbullet=1001=100 m s-1
vball=201=20 m s-1
From the conservation of momentum, Total initial momentum = total final momentum
mbullet×V=mbullet×vbullet+mball×vball
0.01 V=0.01 ×100+0.2×20
V=500 m s-1

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