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183_notes:examples:a_ping-pong_ball_hits_a_stationary_bowling_ball_head-on [2014/11/06 03:50] – 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 | ||
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=== 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 | ||
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Surroundings: | Surroundings: | ||
- | {{183_notes: | + | {{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: |
$$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$$ | $$\vec{p}_{1f} + \vec{p}_{2f} = \vec{p}_{1i} + \vec{p}_{2i}$$ | ||
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$$-\vec{p}_{1i} + \vec{p}_{2f} = \vec{p}_{1i}$$ | $$-\vec{p}_{1i} + \vec{p}_{2f} = \vec{p}_{1i}$$ | ||
+ | |||
+ | Add like terms and rearrange: | ||
$$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | $$\vec{p}_{2f} = 2\vec{p}_{1i}$$ | ||
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[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. |