Question:

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?

Updated On: Sep 3, 2024
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The Correct Option is A

Solution and Explanation

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 : Here, it is assumed that retardation is uniform.
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