Question

A bag of mass $M$ hangs by a long thread and a bullet (mass $m$) comes horizontally with velocity $v$ and gets caught in the bag. Then for the combined system (bag + bullet)

• A. Momentum is $\frac{mMv}{(M+m)}$
  
• B. KE is $\left(\frac{1}{2}\right)M{v}^2$
• C. Momentum is $\frac{mv(M+m)}{M}$
• D. KE is $\frac{m^2{v}^2}{2(M+m)}$

Answer 1

The correct option is B KE is $\left(\frac{1}{2}\right)M{v}^2$

Let V be the velocity of combined system

Conserving the momentum, we get

$\Rightarrow mv=(M+m)V$

$\Rightarrow V=\frac{mv}{M+m}$

Now the K.E of the system = $\frac{1}{2}(M+m)V^2=\frac{m^2{v}^2}{2(M+m)}$

Hence correct answer is option $D$

Also as no external force is applied so the momentum will remain conserved and equal to mv

Hence option $C$ os also correct