Question

If a body loses half of its velocity on penetrating $3cm$ in a wooden block, them how much will it penetrate more before coming to rest ?

• A. $1cm$
• B. $2cm$
• C. $3cm$
• D. $74N$

Answer 1

Answer: (A)

$1cm$

Given,

Initial velocity is $V_{inital}=V$

Velocity at $3cm$ penetration is $V_3=\frac{V}{2}$

Velocity at $(3+x)cm$ is $V_{(3+x)}=0$

Loss in kinetic energy is equal work done by friction

Kinetic Energy = friction force$F_r\times$displacement $d$

At $3cm$ penetration

$\Delta K=F_r\times 3$

$\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\frac{v}{2}\right)}^2=F_r\times 3......\left(1\right)$

At $(3+x)cm$penetration

$\Delta K=F_r\times (3+x)$

$\Rightarrow \frac{1}{2}m{v}^2-0=F_r\times (3+x)......\left(2\right)$

Divide equation $\left(2\right)by\left(1\right)$

$\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\frac{v}{2}\right)}^2}=\frac{F_r\times (3+x)}{F_r\times 3}.$

$\Rightarrow \frac{1}{1-1/4}=\frac{3+x}{3}$

$\Rightarrow x=1cm$