Question

Let the initial velocity of the body be
u. Its velocity becomes u/2 after
penetrating ( 3 mathrm{cm} ) in the block. We
can calculate the deceleration of
the body using these. ( v^{2}=u^{2}+2 a s )
( u^{2} / 4=u^{2}+2 a times 3 )
( -3 u^{2} / 4=6 a )
( a=-u^{2} / 8 )
Now, we can calculate the distance
the body travels till it comes to rest.
( v^{2}=u^{2}+2 a s )
( 0=u^{2}+2 timesleft(-u^{2} / 8right) times s )
( u^{2}=u^{2} s / 4 )
( mathrm{s}=4 mathrm{cm} )
Therefore, the body penetrates ( 1 mathrm{cm} ) ( (4-3=1 mathrm{cm}) ) more before coming to
rest.

# 1.a 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 - 2002, 4/3001 (1) 1 cm (2) 2 cm (D) 4 cm no (3) 3 cm

Solution