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If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? AIEEE  Solved  Paper-2002

A) 1 cm         

B)                           2 cm                         

C)           3 cm                         

D)           4 cm

Correct Answer: A

Solution :

Let initial velocity of body at point A be v, AB is 3 cm.                           From                         \[{{v}^{2}}={{u}^{2}}-2\]as                                 \[{{\left( \frac{v}{2} \right)}^{2}}={{v}^{2}}-2a\times 3\]                                 \[a=\frac{{{v}^{2}}}{8}\]              Let after penetrating 3 cm in a wooden block, the body moves x distance from B to C and comes to rest.              So, for B to C                                 \[u=\frac{v}{2},\,\,v=0\],                                 \[s=x,\,a=\frac{{{v}^{2}}}{8}\]    (retardation)               \[\therefore \]     \[{{(0)}^{2}}={{\left( \frac{V}{2} \right)}^{2}}-2.\frac{{{V}^{2}}}{8}.\,x\]                                 \[x=1\,cm\]              NOTE Here, it is assumed that retardation is uniform.