Question

A bag of mass M hangs by a long massless rope. A bullet of mass m, moving horizontally with velocity u, is caught in the bag. Then for the combined (bag + bullet) system, just after collision :

• A. momentum is muM/(M + m)
• B. kinetic energy is mu${}^2$/2
• C. momentum is mu(M + m)/M
• D. kinetic energy is m${}^2$u${}^2$/2(M + m)

Answer 1

The correct option is D kinetic energy is m${}^2$u${}^2$/2(M + m)

Given : Mass of bag $=M$ $e=0$

Mass of bullet $=m$

Initial velocity of bullet $=u$

Let velocity of "bag + bullet" system after the collision be $V$

Applying conservation of momentum :

$mu+0=(m+M)V$ $⟹V=\frac{mu}{(m+M)}$

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

$K.E=\frac{1}{2}(m+M)\frac{(mu{)}^2}{(m+M{)}^2}$ $⟹K.E=\frac{m^2{u}^2}{2(m+M)}$