Question

In the given diagram a $0.1\text{kg}$ bullet moving with speed $20\text{m/s}$ strikes the $1.9\text{kg}$ mass resting on a smooth block and gets embedded in it. Find the kinetic energy of the mass with which it will strike the ground.

• A. $21\text{J}$
• B. $18\text{J}$
• C. $25\text{J}$
• D. $12\text{J}$

Answer 1

The correct option is A $21\text{J}$

Let $v$ be the speed of mass with it gets embedded in the bullet after collision.
On applying conservation of linear momentum just before and after collision of bullet with mass along horizontal direction(${\overrightarrow{F}}_{\text{ext}}=0$),
$m_1{v}_1+m_2{v}_2=(m_1+m_2)v$
$\Rightarrow 0.1\times 20+1.9\times 0=(0.1+1.9)\times v$
Or, $v=1\text{m/s}$
Now applying work energy theorem on system of bullet and block, just after collision untill just striking the ground :
$W_{\text{ext}}=\Delta K.E$
$W_{\text{ext}}={K.E}_f-{K.E}_i$
Work done by the normal reaction from horizontal surface on system is zero.
$(W_{\text{ext}}=W_N+W_{\text{gravity}}=mgh)$
$\Rightarrow mgh={K.E}_f-\frac{1}{2}m{v}^2$
Here, $m=m_1+m_2$ and when system is about to strike ground the reduction in height is $h=1\text{m}$
$\Rightarrow 2\times 10\times 1=K.E_f-\frac{1}{2}\times 2\times 1^2$
$\Rightarrow {K.E}_f=21\text{J}$