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 $mv$
• D. KE is $\frac{m^2{v}^2}{2(M+m)}$

Answer 1

Answer: (C), (D)

The correct options are
C Momentum is $mv$
D KE is $\frac{m^2{v}^2}{2(M+m)}$

Since, $F_{\text{ext}}=0$
Hence, momentum will remain conserved equal to $mv$.
$mv=(m+M)v^{{}^{\prime }}$
$v^{{}^{\prime }}=\frac{mv}{m+M}$
and
final kinetic energy is $\frac{1}{2}(m+M)v^{\prime 2}=\frac{m^2{v}^2}{2(M+m)}$