Q. If a body loses half of its velocity on penetrating 3cm in a wooden block, then how much will it penetrate more before coming to rest?
JamiaJamia 2005
Solution:
Let initial velocity of body at point A is v, AB is 3 cm.
From $ {{v}^{2}}={{u}^{2}}-2as $ $ {{\left( \frac{v}{2} \right)}^{2}}={{v}^{2}}-2a\times 3 $ $ a=\frac{{{v}^{2}}}{8} $ Let on penetrating 3 cm in a wooden block, the body moves x distance from B to C. So, for B to C $ u=\frac{v}{2},v=0, $ $ s=x,a=\frac{{{v}^{2}}}{8} $ (deceleration) $ \therefore $ $ {{(0)}^{2}}={{\left( \frac{v}{2} \right)}^{2}}-2.\frac{{{v}^{2}}}{8}.x $ $ x=1 $
