Question

A bullet of mass $50\text{gm}$ is fired with a velocity of $90\text{m/s}$ towards a ball of mass $500\text{gm}$, initially at rest. When the bullet strikes the ball, it gets embedded into it and the system starts moving. What is the ratio of the kinetic energy of the bullet to the kinetic energy of the $\text{(bullet + ball)}$ system?

• A. $10.5$
• B. $11$
• C. $8$
• D. $14$

Answer 1

The correct option is B $11$

Initially, $K.E_{\text{bullet}}=\frac{1}{2}\times m\times v^2$
$=\frac{1}{2}\times (50\times {10}^{-3})\times (90{)}^2=202.5\text{J}$
Let velocity of system after collision be $v$.
$m_{\text{system}}=0.05+0.5=0.55\text{kg}$

Applying the principle of conversation of momentum to the system,
$P_i=P_f$
$(0.05\times 90)+0=(0.05+0.5)v$
$\Rightarrow v=\frac{4.5}{0.55}=8.18\text{m/s}$

and Kinetic energy of the system
$K.E_{\text{system}}=\frac{1}{2}\times (0.5+0.05)\times (8.18{)}^2=18.4\text{J}$
Ratio of the kinetic energies:
$\frac{K.E_{\text{bullet}}}{K.E_{\text{system}}}=\frac{202.5}{18.4}=11$