Linear momentum in oblique collisions and generally

[ual8658]

In an oblique collision my understanding is that linear momentum is conserved in all directions (x, y, normal, tangential). But in a constrained oblique collision, does this change?

For example if we had a block lying between two frictionless surfaces with an angled face ( a slope on one face) which a ball going horizontally hits and bounces, the block can only move horizontally but given the normal tangential components, it technically has a velocity component in the normal and tangential directions. However, since the ball hitting the surface shouldn't have an affected tangential velocity component because there is no force of deformation or restoration in the tangential direction, doesn't this mean tangential momentum is not conserved? The block suddenly gained tangential velocity while the ball preserved its own? It would seem that linear momentum horizontally is conserved however.

Edit: After going through this, I see that there must be a force acting on the system if my assumptions are indeed correct. Does this mean that because the sides of the constraint act on the block to prevent movement, linear momentum in the tangential direction is not conserved, and if so, how can one actually explain what happens?

 

[Stephen Tashi]

how can one actually explain what happens?

Perhaps we can explain it by asking "What explains conservation of momentum?"

As I recall, if we begin with F = MA, we can get to F dt = M dv. = d( MV). The basic idea being that when two bodies collide they exert equal and opposite forces on each other over the same time intervals dt. So each body experiences an equal and opposite change in momentum. A complication of modeling real world collisions is that bodies are in contact and moving for finite time intervals during a collision. So the magnitude of the force they exert on each other need not be constant or in the same direction. However, since we have used "infinitestimal" reasoning with dt and d(MV), the argument still convinces a physicist. When force is not constant with respect to time, we have to do an integration to find the net change in momentum.

As I see it, yout question is similar to asking "Is momentum conserved when a falling body hits the ground and lies there?" The standard answer is "Yes, but we must consider the Earth as the body involved in the collision to show that". In you example, the horizontal surfaces and whatever supports them must be considered as bodies involved in the collision.

 

[ual8658]

Perhaps we can explain it by asking "What explains conservation of momentum?"

As I recall, if we begin with F = MA, we can get to F dt = M dv. = d( MV). The basic idea being that when two bodies collide they exert equal and opposite forces on each other over the same time intervals dt. So each body experiences an equal and opposite change in momentum. A complication of modeling real world collisions is that bodies are in contact and moving for finite time intervals during a collision. So the magnitude of the force they exert on each other need not be constant or in the same direction. However, since we have used "infinitestimal" reasoning with dt and d(MV), the argument still convinces a physicist. When force is not constant with respect to time, we have to do an integration to find the net change in momentum.

As I see it, yout question is similar to asking "Is momentum conserved when a falling body hits the ground and lies there?" The standard answer is "Yes, but we must consider the Earth as the body involved in the collision to show that". In you example, the horizontal surfaces and whatever supports them must be considered as bodies involved in the collision.

What do you mean by considering the Earth as a body involved in the collision? Also when considering the coefficient of restitution in these collisions, does that equation relating relative speed of departure and relative speed of arrival not take into consideration conservation of momentum?

 

[A.T.]

What do you mean by considering the Earth as a body involved in the collision?

Your constrain surfaces are attached to the Earth.

Also when considering the coefficient of restitution in these collisions, does that equation relating relative speed of departure and relative speed of arrival not take into consideration conservation of momentum?

No.

 

[ual8658]

Your constrain surfaces are attached to the Earth.No.

Oh ok. But how does that work out mathematically? And so when you say no do you mean the equation with the coefficient of resititution does or does not consider the conservation of angular momentum? Does it work no matter the case?

 

[A.T.]

Oh ok. But how does that work out mathematically? And so when you say no do you mean the equation with the coefficient of resititution does or does not consider the conservation of angular momentum? Does it work no matter the case?

What equation exactly?

 

[ual8658]

What equation exactly?

Coefficient of restitution = (relative speed of departure)/(relative speed of arrival) =
(V_b' - V_a') / (V_a - V_b)

 

[A.T.]

Coefficient of restitution = (relative speed of departure)/(relative speed of arrival) =
(V_b' - V_a') / (V_a - V_b)

This alone says nothing about whether momentum is conserved or not.

 

[ual8658]

This alone says nothing about whether momentum is conserved or not.

Ok thanks, that was my question. But in regards to say a ball bouncing off the Earth and losing height or my example, how do we relate loss of momentum mathematically?