A block of mass 1.9 kg is at rest at the edge of a table, of height 1m. A bullet of mass 0.1kg collides with the block and sticks to it. If the velocity of the bullet is 20m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10m/s2. Assume there is no rotational motion and loss of energy after the collision is negligible]
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a
20J
b
19J
c
23J
d
21J
answer is D.
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Detailed Solution
Conserving momentum , 1.90+0.120= 1.9+0.1v ⇒v = 1m/sThen conserving kinetic energy when it hits the ground=Initial kinetic energy +gain in potential energy=12212+2101K.Efinal=21J
A block of mass 1.9 kg is at rest at the edge of a table, of height 1m. A bullet of mass 0.1kg collides with the block and sticks to it. If the velocity of the bullet is 20m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10m/s2. Assume there is no rotational motion and loss of energy after the collision is negligible]