In a collision where a hand pushes a toy car and we want to find the force using rate of change of momentum the time value is quite intuitive i.e. it's just the time of contact of the hand and the car. However in a collision between a falling block and the ground where there is no rebound how are we supposed to use rate of change of momentum to find the impact force ? After the collision the block just sits on the ground so finding out the time of contact just seems hard to find with no set end point.
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$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Sep 11, 2022 at 16:45
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Collisions are most often a before and after problem. What goes on during a collision is usually not considered.
Collisions occur in a short time and change the momentum and kinetic energy of the colliding objects a lot. That tells you forces are big. When the time is very short, forces can be very big.
In this example, you can't treat the golf ball as a rigid object. Different parts of the ball are subject to different forces and traveling at different velocities. This is not suitable for a classroom physics problem.
On the other hand, before the collision, the ball was round and had velocity $0$. After the collision, the ball is round and has velocity v. You can ask reasonable problems with that.
Here is another post that shows some of the problems with this kind of problem. Ball hits curve of same curvature
answered Sep 11, 2022 at 15:35
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$\begingroup$ Yeah I understand the before and after thought process for collisions and that's why I was asking what the "after" would be in a collision with no rebounds. Because the after situations are indistinguishable. $\endgroup$– PeepachuCommented Sep 11, 2022 at 15:57
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$\begingroup$ If you assume no deformation, you assume the brick instantly comes to a stop, and the contact time is $0$. That is useful for before and after problems. But you have approximated away what happens during the collision. If you don't assume no deformation, you are taking on a harder problem like the golf ball. $\endgroup$ Commented Sep 11, 2022 at 19:19