Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | |||
| 183_notes:examples:a_ping-pong_ball_hits_a_stationary_bowling_ball_head-on [2014/11/06 03:52] – pwirving | 183_notes:examples:a_ping-pong_ball_hits_a_stationary_bowling_ball_head-on [2014/11/06 04:09] (current) – pwirving | ||
|---|---|---|---|
| Line 16: | Line 16: | ||
| === Facts === | === Facts === | ||
| + | |||
| + | Situation occurring in an orbiting spacecraft. | ||
| + | |||
| + | Ping-Pong ball of mass m with initial momentum $\vec{p}_{1i}$ traveling in the +x direction. | ||
| + | |||
| + | Bowling ball of mass M is hit by the Ping-Pong ball while stationary. | ||
| Initial situation: Just before collision | Initial situation: Just before collision | ||
| Line 54: | Line 60: | ||
| === Solution === | === Solution === | ||
| - | From the momentum principle | + | From the momentum principle: |
| $$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$$ | $$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$$ | ||
| Line 61: | Line 67: | ||
| $$-\vec{p}_{1i} + \vec{p}_{2f} = \vec{p}_{1i}$$ | $$-\vec{p}_{1i} + \vec{p}_{2f} = \vec{p}_{1i}$$ | ||
| + | |||
| + | Add like terms and rearrange: | ||
| $$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | $$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | ||
| Line 74: | Line 82: | ||
| [b] Final speed of bowling ball: | [b] Final speed of bowling ball: | ||
| - | $$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M} | + | From the equation for momentum: $p_{2f} = M(v_{2f})$ |
| + | |||
| + | Therefore: | ||
| + | |||
| + | $$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M}$$ | ||
| + | |||
| + | Substitute $2\vec{p}_{1i}$ in for ${p_{2f}}$ from previous result in this example above. | ||
| + | |||
| + | $$\vec{v}_{2f} \approx | ||
| + | |||
| + | Momentum is mass times velocity so substitute this in: | ||
| + | |||
| + | $$\vec{v}_{2f} \approx | ||
| + | |||
| + | Rearrange: | ||
| + | |||
| + | $$\vec{v}_{2f} \approx | ||
| This is a very small speed since m<<M/ For example, if the mass of the bowling ball is about 5 kg, and the gram Ping-Pong ball is initially traveling at 10 m/s, the final speed of the bowling ball will be 0.008 m/s. | This is a very small speed since m<<M/ For example, if the mass of the bowling ball is about 5 kg, and the gram Ping-Pong ball is initially traveling at 10 m/s, the final speed of the bowling ball will be 0.008 m/s. | ||
| [c] Kinetic energies: | [c] Kinetic energies: | ||
| + | |||
| + | From our representations we know that $K = \frac{1}{2}m(\frac{p^{2}}{m})$ | ||
| + | |||
| + | But we also know that $p_{2f} = 2\vec{p}_{1i}$ from earlier so substituting this into $K = \frac{1}{2}m(\frac{p^{2}}{m})$ we get this equation for the final kinetic energy of the bowling ball | ||
| $$K_{2f} = \dfrac{(2p_{1i})^2}{2M}$$ | $$K_{2f} = \dfrac{(2p_{1i})^2}{2M}$$ | ||
| - | and | + | and the equation for the final kinetic energy of the Ping-Pong ball is |
| $$K_{1f} = \dfrac{p^2_{1i}}{2m}$$ | $$K_{1f} = \dfrac{p^2_{1i}}{2m}$$ | ||
| + | |||
| + | As we assumed that the speed of the Ping-Pong ball does not change significantly after the collision. | ||
| Because the mass of the bowling ball is much larger than the mass of the Ping-Pong ball, the kinetic energy of the bowling ball is much smaller than the kinetic energy of the Ping-Pong ball. The kinetic energy of the 2g Ping-Pong ball, traveling at 10m/s, is about 0.1J, while the 5kg bowling ball has acquired a kinetic energy of $1.6 x 10^{-4} J$ - nearly 1000 times less. | Because the mass of the bowling ball is much larger than the mass of the Ping-Pong ball, the kinetic energy of the bowling ball is much smaller than the kinetic energy of the Ping-Pong ball. The kinetic energy of the 2g Ping-Pong ball, traveling at 10m/s, is about 0.1J, while the 5kg bowling ball has acquired a kinetic energy of $1.6 x 10^{-4} J$ - nearly 1000 times less. | ||