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| 183_notes:examples:maximally_inelastic_collision_of_two_identical_carts [2014/11/04 07:17] – pwirving | 183_notes:examples:maximally_inelastic_collision_of_two_identical_carts [2014/11/06 03:46] (current) – pwirving | ||
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| === Lacking === | === Lacking === | ||
| + | Find the final momentum, final speed, and final kinetic energy of the carts in terms of their initial values. | ||
| + | What is the change in internal energy of the two carts? | ||
| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
| - | Neglect | + | External forces are negligible during the collision, so neglect |
| === Representations === | === Representations === | ||
| Line 30: | Line 33: | ||
| Surroundings: | Surroundings: | ||
| + | {{183_notes: | ||
| + | |||
| + | $\vec{p}_f = \vec{p}_i + \vec{F}_{net} \Delta t$ | ||
| + | |||
| + | $E_f = E_i + W + Q$ | ||
| + | |||
| + | $K = \frac{1}{2}mv^{2} = \frac{1}{2}m(\frac{p}{m})^{2} = \frac{1}{2}m(\frac{p^{2}}{m})$ | ||
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| Since the y and z components of momentum don't change, we can work with only x components | Since the y and z components of momentum don't change, we can work with only x components | ||
| - | From the momentum principle (x components): | + | From the momentum principle (x components) |
| $${p}_{1xf} + {p}_{2xf} = {p}_{1xi}$$ | $${p}_{1xf} + {p}_{2xf} = {p}_{1xi}$$ | ||
| + | |||
| + | After the collision ${p}_{2xf}$ is equal to ${p}_{1xf}$ as they are stuck together so: | ||
| $$2p_{1xf} = p_{1xi}$$ | $$2p_{1xf} = p_{1xi}$$ | ||
| + | |||
| + | Rearrange to isolate $p_{1xf}$ | ||
| $$p_{1xf} = \dfrac{1}{2}p_{1xi}$$ | $$p_{1xf} = \dfrac{1}{2}p_{1xi}$$ | ||
| - | The final speed of the stuck-together carts its half the initial speed: | + | Therefore the final speed of the stuck-together carts its half the initial speed: |
| $$v_{f} = \dfrac{1}{2}{v_{i}}$$ | $$v_{f} = \dfrac{1}{2}{v_{i}}$$ | ||
| - | Final translational kinetic energy: | + | Since we know the speed of the carts we can calculate their translational kinetic energy. |
| + | |||
| + | Final translational kinetic energy | ||
| $$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}mv^2_{f})$$ | $$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}mv^2_{f})$$ | ||
| + | |||
| + | Substitute in the final speed of the stuck-together carts: | ||
| $$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}m(\dfrac{1}{2}v_{i})^2) = \dfrac{1}{4}mv^2_{i}$$ | $$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}m(\dfrac{1}{2}v_{i})^2) = \dfrac{1}{4}mv^2_{i}$$ | ||
| + | |||
| + | Substituting back in $K_{1i}$ in order to put the final kinetic energy in terms of the initial kinetic energy we get: | ||
| $$(K_{1f} + K_{2f}) = \dfrac{K_{1i}}{2}$$ | $$(K_{1f} + K_{2f}) = \dfrac{K_{1i}}{2}$$ | ||
| - | From the energy principle: | + | We know the initial and final translational kinetic energies of the system, so we can use the energy principle |
| $$K_{1f} + K_{2f} + E_{int,f} = K_{1i} + E_{int,i}$$ | $$K_{1f} + K_{2f} + E_{int,f} = K_{1i} + E_{int,i}$$ | ||
| + | |||
| + | Rearrange to get the final internal energy minus the initial internal energy one one side. | ||
| $$E_{int,f} - E_{int,i} = K_{1i} - (K_{1f} + K_{2f})$$ | $$E_{int,f} - E_{int,i} = K_{1i} - (K_{1f} + K_{2f})$$ | ||
| + | |||
| + | $E_{int,f} - E_{int,i}$ is the same as $\Delta E_{int}$ and this is what we are trying to find so substitute this in. Also substitute $\dfrac{K_{1i}}{2}$ for $(K_{1f} + K_{2f})$. We get: | ||
| $$\Delta E_{int} = K_{1i} - \dfrac{K_{1i}}{2}$$ | $$\Delta E_{int} = K_{1i} - \dfrac{K_{1i}}{2}$$ | ||
| + | |||
| + | Resolve the right hand side and you get: | ||
| $$\Delta E_{int} = \dfrac{K_{1i}}{2}$$ | $$\Delta E_{int} = \dfrac{K_{1i}}{2}$$ | ||