# Thread: Newton's 3rd Law

1. If an object (for example a bowling ball) is throw, and begins slowing down due to friction (there is no propulsive force accelerating it), hits a pin, according to Newton's 3rd Law a force is exerted on to the pin and the bowling ball in the opposite direction. Where does this forwards force come from?

Is impact force and force that causes acceleration different?

Does it have something to do with Work = Force x Displacement?  2.

3. At time 0 you apply a force to the ball that gets it moving. At time I the ball hits the pins. In between, the lane is applying a friction force against the ball accelerating it against the direction of motion. (All change in velocity is acceleration.) When the ball hits the pin, both apply a force to each other. The same magnitude of force in opposite directions. The difference is that the ball is much heavier and so accelerates less from that force. Generally, that acceleration isn't enough to stop or reverse the ball, so it continues in the same general direction, while the pin goes from stopped to flying away.

None of this really has anything to do with work. You could calculate how much work you did to the ball, or how much work the ball did to the pins, but that wouldn't shed any light on anything in this scenario.  4. Originally Posted by jeffreyquach Does it have something to do with Work = Force x Displacement?
Work = Force x Displacement does apply, but it's explained better by Force = Mass x Acceleration. When the ball collides with the pin, some acceleration is going to happen because the ball and the pin cannot keep going the same speed as they were before the collision while occupying the same space. For acceleration to happen, there must be a force.

It is true that Work = Force x Displacement but this formula is more useful if you have a constant force applied over a measurable distance. Collisions happen fast, the forces change fast, and distances are short. To analyze collisions, you generally use conservation of momentum and if applicable, conservation of kinetic energy. Momentum is always conserved in a collision. Kinetic energy is conserved in an elastic collision. The bowling ball colliding with the pins is going to me mostly elastic. An inelastic collision would be one where the objects stay together after the collision instead of bouncing away from each other.  5. Ohh I see. Thank you for clarifying this.

Is there a mathematical way which shows that the slower an object is travelling, the less force it will exert upon colliding?

Would it make sense if i said:

Momentum = mass x velocity

Therefore, the slower an object the less momentum it will have. Upon collision with another object, there will be a change in momentum.

But the change in momentum is impulse, which equals Force x Change in time

Therefore with less velocity, there is less momentum which results in less change in momentum which can be attributed to a smaller force?  6. Originally Posted by jeffreyquach Ohh I see. Thank you for clarifying this.

Is there a mathematical way which shows that the slower an object is travelling, the less force it will exert upon colliding?

Would it make sense if i said:

Momentum = mass x velocity

Therefore, the slower an object the less momentum it will have. Upon collision with another object, there will be a change in momentum.

But the change in momentum is impulse, which equals Force x Change in time

Therefore with less velocity, there is less momentum which results in less change in momentum which can be attributed to a smaller force?
Yes I think impulse is a helpful concept here. This is Force x time for which it is exerted and results in a given change in momentum With a rapidly moving object the rebound takes place more quickly so the force is bigger but is exerted for a shorter time than for a slower moving object. The total impulse imparted is the same and so the total change of momentum is the same but the rate of change is different.  7. Originally Posted by exchemist  Originally Posted by jeffreyquach Ohh I see. Thank you for clarifying this.

Is there a mathematical way which shows that the slower an object is travelling, the less force it will exert upon colliding?

Would it make sense if i said:

Momentum = mass x velocity

Therefore, the slower an object the less momentum it will have. Upon collision with another object, there will be a change in momentum.

But the change in momentum is impulse, which equals Force x Change in time

Therefore with less velocity, there is less momentum which results in less change in momentum which can be attributed to a smaller force?
Yes I think impulse is a helpful concept here. This is Force x time for which it is exerted and results in a given change in momentum With a rapidly moving object the rebound takes place more quickly so the force is bigger but is exerted for a shorter time than for a slower moving object. The total impulse imparted is the same and so the total change of momentum is the same but the rate of change is different.
I would have gone with the kinetic energy which is one half mass times velocity squared. It equals work.
Something travellig 2 times as fast as somthing else of the same weight will take 4 times as far to stop, it will also do 4 times as much damage if it hits something.  8. let's say there is one object moving towards a second one with uniform motion. after a time the objects collide and the second object does hardly move at all. what are the forces that the objects experience??... according to newtons 3rd law the forces should be equal in size and opposite in direction... and according to newton's second law F=ma=m*dv/dt=dp/dt. during collision, the first object decelerates instantly. this means that the momentum "jumps" from some value to zero, which makes it impossible to take its derivative and therefore impossible to calculate the Force at that moment ...but how is it then possible to calculate the forces... please correct any mistakes in my thinking...  9. Originally Posted by bilbaolove let's say there is one object moving towards a second one with uniform motion. after a time the objects collide and the second object does hardly move at all. what are the forces that the objects experience??... according to newtons 3rd law the forces should be equal in size and opposite in direction... and according to newton's second law F=ma=m*dv/dt=dp/dt. during collision, the first object decelerates instantly. this means that the momentum "jumps" from some value to zero, which makes it impossible to take its derivative and therefore impossible to calculate the Force at that moment ...but how is it then possible to calculate the forces... please correct any mistakes in my thinking...
There is no such thing as instant deceleration , so . Momentum cannot "jump". All these things are idealizations of what happens in reality.  10. Originally Posted by bilbaolove let's say there is one object moving towards a second one with uniform motion. after a time the objects collide and the second object does hardly move at all. what are the forces that the objects experience??... according to newtons 3rd law the forces should be equal in size and opposite in direction... and according to newton's second law F=ma=m*dv/dt=dp/dt. during collision, the first object decelerates instantly. this means that the momentum "jumps" from some value to zero, which makes it impossible to take its derivative and therefore impossible to calculate the Force at that moment ...but how is it then possible to calculate the forces... please correct any mistakes in my thinking...
No, it isn't instantaneous because all materials have some elasticity.  11. Originally Posted by bilbaolove let's say there is one object moving towards a second one with uniform motion. after a time the objects collide and the second object does hardly move at all. what are the forces that the objects experience??... according to newtons 3rd law the forces should be equal in size and opposite in direction... and according to newton's second law F=ma=m*dv/dt=dp/dt. during collision, the first object decelerates instantly. this means that the momentum "jumps" from some value to zero, which makes it impossible to take its derivative and therefore impossible to calculate the Force at that moment ...but how is it then possible to calculate the forces... please correct any mistakes in my thinking...
Yes, to expand on what the other have said, you are implicitly using a simplifying assumption here, which is that the bodies are perfectly rigid. This means you get divisions by zero if you try to work out the rates of change during the collision event - hence you can't do it. (It is equivalent to having an infinite force applying for an infinitely small period of time.) In reality, any bodies in collision progressively deform, which makes the process take a finite time, hence no division by zero and you have a problem you can in principle solve.

Once you have a finite time for the interaction then you can in principle try to work out the force from the impulse (F x t) during the interaction, which is equal to the change in momentum. But it will be complicated because the force will rise progressively from zero to a maximum as the objects deform and then diminish as they begin to rebound, in a way that will depend on the materials and the shape of the parts of the object in contact.  Bookmarks
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