Question

When a gun of mass $M$ fires a bullet of mass $m$, the total energy released in the explosion is found to be $E$. The kinetic energy of the bullet is:

• A. $\frac{EM}{(M+m)}$
• B. $\frac{Em}{(M+m)}$
• C. $\frac{E(M+m)}{M}$
• D. $\frac{E(M+m)}{m}$

Answer 1

Answer: (A)

$\frac{EM}{(M+m)}$

Using momentum conservation on bullet + gun as a system

$\Rightarrow$ initial momentum$=$ Final momentum

$0=-MV+mv$

$\Rightarrow mv=MV$..........(1)

Now total energy $=E$

$\Rightarrow \frac{1}{2}M{V}^2+\frac{1}{2}m{v}^2=E$

From (1)

$V=\frac{mv}{M}$

$\Rightarrow \frac{1}{2}M\frac{m^2}{M^2}{v}^2+\frac{1}{2}m{v}^2=E$

$\frac{1}{2}m{v}^2(\frac{m}{M}+1)=E$

$\frac{1}{2}m{v}^2=\frac{EM}{M+m}$

$\downarrow$

K.E bullet