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Q.
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?
AIEEEAIEEE 2002Motion in a Straight Line
Solution:
Let initial velocity of body at point A is v, AB is 3 cm.
From
$\nu^{2}=u^{2}-2 a s$
$\left(\frac{v}{2}\right)^{2} =v^{2} \cdot 2 a \times 3$
$a =\frac{v^{2}}{8}$
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)
$\therefore (0)^{2} =\left(\frac{v}{2}\right)^{2}-2 \cdot \frac{v^{2}}{8} \cdot x$
$x =1$ cm Note: Here, it is assumed that retardation is uniform.
