Question

A 50 gram bullet moving a velocity of mass 10 $m{s}^{-1}$ embedded into a 950 g stationary body. The loss in K.E. the system will be:

• A. 95%
• B. 100%
• C. 5%
• D. 50%

Answer 1

The correct option is C 5%

This 0 is an inelastic collision (c = 0)

The loss of KE due to collision is given by:-

$\Delta K=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(u_2-u_1{)}^2$

Put value ,$u_1=10m/s$ and $u_2=0$

$\Rightarrow \Delta K=\frac{1}{2}\times \frac{950\times 10}{950+10}\times (0-10{)}^2$

$=\frac{1}{2}\times \frac{9500}{1000}\times 100=475mJ$

also initial energy of system :-

$k=\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}\times 50\times {10}^2=2500m/s$

$\therefore$ loss in KE,

$\frac{\Delta K}{K}=\frac{475}{2500}=0.0475=4.75\%$