Question

A $50gm$ bullet moving with a velocity of $10m/s$ gets embedded into a $950gm$ stationary body. The loss in kinetic energy of the system will be:

• A. $5\%$
• B. $50\%$
• C. $100\%$
• D. $95\%$

Answer 1

The correct option is D $95\%$

$\text{Given:}$For bullet, $m=0.050kg$, $v_i=10m/s$

For body, $M=0.950kg$, $V_i=0$

$\text{Step 1: Apply conservation of momentum}$

Let $V_f$ be the common velocity of both after collision.

There is no external force acting on the system therefore momentum is conserved.

$P_i=P_f$

$\Rightarrow m{v}_i+M{V}_i=(m+M)V_f$

$\Rightarrow (0.050\times 10)+(0.950\times 0)=(0.050+0.950)V_f$

$\Rightarrow V_f=0.5m/s$

$\text{Step 2: Change in Kinetic energy}$

Initial kinetic energy, $K.E_i=\frac{1}{2}m{v_i}^2+\frac{1}{2}M{V}_i^2$

$=\frac{1}{2}\times 0.050\times {10}^2+\frac{1}{2}\times 0.950\times 0^2=2.5J$

Final kinetic energy, $K.E_f=\frac{1}{2}(m+M){V_f}^2$

$=\frac{1}{2}\times (0.050+0.950)\times {0.5}^2=0.125J$

Therefore, Change in Kinetic energy, $\Delta K.E=K.E_i-K.E_f$

$=2.5-0.125=2.375J$

$\text{Step 3: Percentage loss of Kinetic energy}$

Percentage loss of $K.E=\frac{\Delta KE}{K{E}_i}\times 100\%$

$=\frac{2.375}{2.5}\times 100\%=95\%$

Hence Option (D) is correct.