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| 183_notes:examples:a_ping-pong_ball_hits_a_stationary_bowling_ball_head-on [2014/11/04 07:55] – pwirving | 183_notes:examples:a_ping-pong_ball_hits_a_stationary_bowling_ball_head-on [2014/11/06 04:09] (current) – pwirving | ||
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| ===== Example: A Ping-Pong Ball Hits a Stationary Bowling Ball Head-on ===== | ===== Example: A Ping-Pong Ball Hits a Stationary Bowling Ball Head-on ===== | ||
| - | In an orbiting spacecraft a Ping-Pong ball of mass m (object 1) traveling in the +x direction with initial momentum \vec{p}_1i hits a stationary bowling ball of mass M (object 2) head on, as shown in the figure in representations. | + | In an orbiting spacecraft a Ping-Pong ball of mass m (object 1) traveling in the +x direction with initial momentum |
| What are the | What are the | ||
| - | (a) momentum? | + | [a] momentum? |
| - | (b) speed? | + | [b] speed? |
| - | (c) kinetic energy? | + | [c] kinetic energy? |
| Of each object after the collision. | Of each object after the collision. | ||
| 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 24: | Line 30: | ||
| === Lacking === | === Lacking === | ||
| + | What are the | ||
| + | [a] momentum? | ||
| + | |||
| + | [b] speed? | ||
| + | |||
| + | [c] kinetic energy? | ||
| + | |||
| + | Of each object after the collision. | ||
| Line 37: | Line 51: | ||
| Surroundings: | Surroundings: | ||
| + | {{183_notes: | ||
| + | |||
| + | $\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$ | ||
| + | |||
| + | $K = \frac{1}{2}m(\frac{p^{2}}{m})$ | ||
| === 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 48: | 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: | ||
| $$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | $$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | ||
| - | (a) The final momentum of the bowling ball is twice the initial momentum of the Ping-Pong ball. | + | [a] The final momentum of the bowling ball is twice the initial momentum of the Ping-Pong ball. |
| It may be surprising that the bowling ball ends up with about twice the momentum of the Ping-Pong ball. One way to understand this is that the final momentum | It may be surprising that the bowling ball ends up with about twice the momentum of the Ping-Pong ball. One way to understand this is that the final momentum | ||
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| The Ping-Pong ball's speed hardly changed, but its momentum changed a great deal. Because momentum is a vector, a change of direction is just as much a change of magnitude. This big change is of course due to the interatomic electric contact forces exerted on the Ping-Pong ball by the bowling ball. By reciprocity, | The Ping-Pong ball's speed hardly changed, but its momentum changed a great deal. Because momentum is a vector, a change of direction is just as much a change of magnitude. This big change is of course due to the interatomic electric contact forces exerted on the Ping-Pong ball by the bowling ball. By reciprocity, | ||
| - | (b) Final speed of bowling ball: | + | [b] Final speed of bowling ball: |
| + | |||
| + | 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. | ||
| + | |||
| + | $$\vec{v}_{2f} \approx \dfrac{2p_{1i}}{M}$$ | ||
| + | |||
| + | Momentum is mass times velocity so substitute this in: | ||
| + | |||
| + | $$\vec{v}_{2f} \approx \dfrac{2mv_{1i}}{M}$$ | ||
| + | |||
| + | Rearrange: | ||
| + | |||
| + | $$\vec{v}_{2f} \approx \dfrac{m}{M})v_{1i}$$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | 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 | ||
| + | |||
| + | $$K_{2f} = \dfrac{(2p_{1i})^2}{2M}$$ | ||
| + | |||
| + | and the equation for the final kinetic energy of the Ping-Pong ball is | ||
| + | |||
| + | $$K_{1f} = \dfrac{p^2_{1i}}{2m}$$ | ||
| + | |||
| + | 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. | ||
| - | $$\vec{v}_{2f} \approx \dfrac{p_{2f}}{M} = \dfrac{2p_{1i}}{M} = \dfrac{2mv_{1i}}{M}$$ = 2\dfrac({m}{M})v_{1i}$$ | ||