Question 9.54: A completely inelastic collision occurs between two balls of......
A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 20 m/s and the other ball, of mass 2.0 kg, is moving downward at 12 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)
The total momentum immediately before the collision (with +x upward) is
p_{i}=(3.0\ k{g})(20\ {m}/s)+(2.0\ k{g})(-12\ {m}/s)=36\mathrm{~kg}.{m}/s\,.
Their momentum immediately after, when they constitute a combined mass of M = 5.0 kg, is p_{f}=(5.0\ k{g}){\vec{\nu}}. By conservation of momentum, then, we obtain \vec{\nu} = 7.2 m/s, which becomes their “initial” velocity for their subsequent free-fall motion. We can use Ch. 2 methods or energy methods to analyze this subsequent motion; we choose the latter. The level of their collision provides the reference (y = 0) position for the gravitational potential energy, and we obtain
K_{0}+U_{0}\ =\ K+U\ \ \Rightarrow\ \ \frac{1}{2}M\nu_{0}^{2}+0\ =\ 0+M gy_{\mathrm{max}}\,.
Thus, with \nu_{0}=7.2\;\mathrm{m}/s, we find y_{\mathrm{max}}=2.6\ \mathrm{m}.