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I have a stationary cart of varying mass placed at a set location on a linear track. I have another cart, the incident cart which has a wound up spring to launch it from a wall on the same linear track.

When I release the spring, the incident cart goes ahead and collides with the stationary cart, causing then the stationary cart to move.

A few questions relating to this phenomenon:

  • Is this collision inelastic or elastic? I'm thinking it is elastic because of the perfect transfer of energy.

  • How can I explain that when the mass goes up the velocity of the stationary cart after collision goes down (that relates to the conservation of momentum)?

Thanks

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  • $\begingroup$ What do you think? About what the collision is? $\endgroup$
    – Kunal Pawar
    Commented Apr 16, 2017 at 15:32
  • $\begingroup$ @KunalPawar I think the collision is a perfectly elastic collision under the perfect circumstances due to the perfect transfer of energy from the incident cart to the stationary cart. Also, the carts do not deform (they do but in negligible amounts) during the collision. $\endgroup$
    – Roger Ho
    Commented Apr 16, 2017 at 15:37

1 Answer 1

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In reality, you can never have a 100% elastic collision. There's always energy dissipated in the form of heat or used in the distortion of the colliding bodies (think of a tennis ball in slow motion: it's distorted in the moment of impact and then goes back to its original shape). But, in this case, as the dissipated energy might be infinitesimal, I believe you can make the approximation that the energy is conserved.

Regarding the second question, there is conservation of momentum: the product of mass times the initial velocity of the incident cart ($\ p_{i1}=m_1v_{i1}$) equals the sum of mass times final velocity of each cart ($\ p_{f1} + p_{f2}= m_1v_{f1}+m_2v_{f2}$). So, if cart 2 has a bigger mass (m2), in order to mantain the value of $\ p_{f1} + p_{f2} = p_{i1}$, the final velocity of the stationary cart (vf2) has to be smaller.

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  • $\begingroup$ Thank you for your reply. A quick question that extends on kinetic energy and momentum. Would it be ok the assume that because 2 objects have the same magnitude of kinetic energy acting on them they would have the same momentum? Is there a more mathematical way / more logical way to infer this? $\endgroup$
    – Roger Ho
    Commented Apr 16, 2017 at 16:11
  • $\begingroup$ linear momentum is the derivate of kinetic energy in order to velocity (in a non-relativistic scenario) $\endgroup$
    – Jose Carvalho
    Commented Apr 16, 2017 at 17:00
  • $\begingroup$ If they have the same momentum, that does not imply that they have the same energy and vice versa $\endgroup$
    – Jose Carvalho
    Commented Apr 16, 2017 at 17:03

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