Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

107 of 130

Question 107 of 130

A block of mass 1.9kg1.9kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1kg0.1kg collides with the block and sticks to it. If the velocity of the bullet is 20ms−120m{s}^{-1} in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g=10ms−2g=10m{s}^{-2}. Assume there is no rotational motion and loss of energy after the collision is negligible.]

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Answer: (A) 21J21J

Conservation of linear momentum
0.1×20=(0.1+1.9)×v0.1\times 20=\left(0.1+1.9\right)\times v
v=1ms−1v=1m{s}^{-1}
Using work energy theorem
Wg=∆k{W}_{g}=∆k
2×g×1=k−12×2×122\times g\times 1=k-\dfrac{1}{2}\times 2\times {1}^{2}
∴k=21J∴k=21J

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