183_notes:examples:elastic_collision_of_two_identical_carts

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Cart 1 collides with stationary cart 2, which is identical. Suppose that the collision is (nearly) elastic, as it will be if the carts repel each other magnetically or interact through soft springs. In this case there is no change of internal energy. What are the final momenta of the two carts?

Facts

Initial situation: Just before collision

Final situation: Just after collision

Lacking

Final momenta of the two carts

Approximations & Assumptions

Assume there is no change of internal energy

Neglect friction and air resistance

Representations

System: Both carts

Surroundings: Earth, track, air

Solution

Since the y and z components of momentum don't change, we can work with only x components

From the momentum principle:

$$\vec{p}_f = \vec{p}_i + \vec{F}_{net} \Delta t$$

$$\vec{p}_{1xf} + \vec{p}_{2xf} = \vec{p}_{1xi} + 0$$

From the energy principle:

$$E_f = E_i + W + Q$$

$$K_{1f} + K_{2f} + E_{int1f} + E_{int2f} = K_{1i} + K_{2i} + E_{int1i} + E_{int2i}$$

$$K_{1f} + K_{2f} = K_{1i}$$

Combine momentum and energy equations:

$$\dfrac{p^{2}_{1xf}}{2m} + \dfrac{p^{2}_{2xf}}{2m} = \dfrac{(p_{1xf} + p_{2xf})^2}{2m}$$

$$p^{2}_{1xf} + p^{2}_{2xf} = p^{2}_{1xf} + 2{p_{1xf}p_{2xf}}$$ + p^{2}_{2xf}$$

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  • Last modified: 2014/11/04 06:45
  • by pwirving