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A 1.9 kg block rests at the edge of a 1 m high table. A 0.1 kg bullet moving horizontally at 20 m/s strikes the block and embeds in it. Neglecting rotation and any energy loss after the collision, find the kinetic energy of the block-plus-bullet system just before it hits the floor. (g = 10 m/s²)
- 21 J
- 23 J
- 19 J
- 20 J
Correct answer: 21 J
Solution
Momentum conservation gives a post-collision speed of 1 m/s, so KE just after collision is 1 J; falling 1 m adds mgh = 20 J, giving 21 J just before impact.
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