Question 9.57: In Fig. 9-61, a ball of mass m = 60 g is shot with speed νi ......
In Fig. 9-61, a ball of mass m = 60 g is shot with speed \nu_{i} = 22 m/s into the barrel of a spring gun of mass M = 240 g initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after the ball stops in the barrel? (b) What fraction of the initial kinetic energy of the ball is stored in the spring?
(a) Let ν be the final velocity of the ball-gun system. Since the total momentum of the system is conserved m\nu_{i} = (m + M)ν. Therefore,
\nu={\frac{m\nu_{i}}{m+M}}={\frac{(60\ 8)(22\mathrm{~m/s})}{60\mathrm{~g}+240\mathrm{~g}}}=4.4\mathrm{~m/s}\,.
(b) The initial kinetic energy is K_{i}={\frac{1}{2}}m \nu_{i}^{2} and the final kinetic energy is
K_{f}={\textstyle{\frac{1}{2}}}\bigl(m+M\bigr)\nu^{2}={{\frac{1}{2}}}m^{2}\nu_{i}^{2}/\bigl(m+M\bigr)\,.
The problem indicates \Delta E_{\mathrm{th}} = 0 , so the difference K_{i}-K_{f} must equal the energy U_{s} stored in the spring:
U_{s}=\frac{1}{2}m\nu_{i}^{2}-\frac{1}{2}\frac{m^{2}\nu_{i}^{2}}{\left(m+M\right)}=\frac{1}{2}m\nu_{i}^{2}\left(1-\frac{m}{m+M}\right)=\frac{1}{2}m\nu_{i}^{2}\,\frac{M}{m+M}.
Consequently, the fraction of the initial kinetic energy that becomes stored in the spring is
{\frac{U_{s}}{K_{i}}}={\frac{M}{m+M}}={\frac{240}{60+240}}=0.80\,.