Chapter 10, Problem 45PQ

A model rocket is shot straight up and explodes at the top of its trajectory into three pieces as viewed from above and shown in Figure P10.44. The masses of the three pieces are m_A = 100.0 g, m_B = 20.0 g, and m_C = 30.0 g. Immediately after the explosion, piece A is traveling at 1.50 m/s, and piece B is traveling at 7.00 m/s in a direction 30° below the negative x axis as shown. What is the velocity of piece C?

FIGURE P10.44

Problems 44 and 45.

45. We can use the conservation of momentum (Eq. 10.9). The total initial momentum is zero, so the sum of all the final momenta should be zero.

$m_{\text{A}}{\stackrel{\to }{v}}_{\text{A}f}+m_{\text{B}}{\stackrel{\to }{v}}_{\text{B}f}+m_{\text{C}}{\stackrel{\to }{v}}_{\text{C}f}=0$

This velocities for A and B can be expressed as vectors.

$\begin{array}{l}{\stackrel{\to }{v}}_{\text{A}f}=1.50\hat{j}\text{ m/s}\\ {\stackrel{\to }{v}}_{\text{B}f}=-(7.00\hat{i}\text{ m/s})\cos 30°-(7.00\hat{j}\text{ m/s})\sin 30°=(-6.06\hat{i}-3.50\hat{j})m\text{/s}\end{array}$

We can now solve the momentum equation.

$\begin{array}{l}(100.0\text{ g})(1.50\hat{j}\text{ m/s})+(20.0\text{ g})(-6.06\hat{i}-3.50\hat{j})\text{m/s}+(30.0\text{ g}){\stackrel{\to }{v}}_{\text{C}f}=0\\ {\stackrel{\to }{v}}_{\text{C}f}=\boxed{(4.04\hat{i}-2.67\hat{j})\text{m/s}}\end{array}$

The velocity of piece C is down and to the right as expected.