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Q.

A block of mass 1.9 kg is rest at the edge of a table of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision, then the kinetic energy just before the combined system strikes the floor, is (Take, g=10 m/s2) and assume there is no rotational motion and loss of energy after the collision is negligible)

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a

21 J

b

23 J

c

19 J

d

20 J

answer is C.

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Detailed Solution

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Perfectly inelastic collision  By applying L.C.M

Question Image

mu=(m+M)v

v=mum+M=0.1×20(0.1+1.9)=1m/s

Total energy of bullt and block after collision

TE=PE+KE

12(m+M)V2+(m+M)gh

=1+20=21J

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