Question

A bag P (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 (bag+bullet) system

• A. momentum is $\frac{mvM}{M+m}$.
• B. kinetic energy is $\frac{m{V}^2}{2}$.
• C. momentum is $\frac{mv(M+m)}{M}$.
• D. kinetic energy is $\frac{m^2{v}^2}{2(M+m)}$.

Answer 1

Answer: (D)

kinetic energy is $\frac{m^2{v}^2}{2(M+m)}$.

The combined velocity of the bag+bullet system, $v_f=m\frac{v}{m+M}$
Now, we observe that the options talk about momentum and kinetic energy, so we will calculate the value of these quantities one by one:
(i) Momentum = (m+M)$v_f$=mv
Note that, the final momentum must equal the initial momentum, so this value of final momentum could have been directly written.
(ii)Kinetic energy $=\frac{1}{2}(m+M){v_f}^2$

$=\frac{1}{2}(m+M){\left[m\frac{v}{m+M}\right]}^2$

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

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