Question

A $20$g bullet pierces through a plane of mass $M_1=1kg$ and then comes to rest inside a second plate of mass $M_2=2.98kg$ as shown in Fig. It is found that the two plates, initially at rest, now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between $M_1$ and $M_2$. Neglect any loss of material of the plates due to the action of bullet.

• A. $50$
• B. $25$
• C. $100$
• D. $75$

Answer 1

The correct option is B $25$

$\frac{1}{2}m{u}^2=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}m{u}_1^2\to \left(1\right)$

$\Rightarrow \frac{1}{2}m{u}_1^2=\frac{1}{2}(m_2+m)v^2$

$\Rightarrow \frac{1}{2}20u_1^2=\frac{1}{2}[2980+20]v^2$

$\Rightarrow \frac{20m^2}{2}=\frac{v^{2}3000}{2}$

$\Rightarrow u_1^2=150v^2\to \left(2\right)$ $u_1=12.24T$

$\Rightarrow \frac{1}{2}20u^2=\frac{1}{2}1000v^2+\frac{1}{2}20\times 150v^2$

$\Rightarrow 20u^2=[1000+3000]v^2$

$\Rightarrow 20u^2=4000v^2$

$\Rightarrow u^2=\frac{4000}{20}{v}^2$

$\Rightarrow u^2=200v^2$ $14.14v=u$

$=\frac{u_1-u}{u}\times 100$

$=\frac{14.14-12.247}{14.14}\simeq 15$

Hence, the answer is $15.$