How much work done by a ball during an oblique collision?

[sfensphan]

Suppose I have a ball colliding with a surface at an oblique angle. I want to know what work will be done on the surface. Now, there is kinetic energy associated with the tangential and normal velocity of the ball. Can I say that the kinetic energy of the ball associated with the normal velocity is the only energy that does work on the surface? That makes sense to me, but I read that kinetic energy is a scalar, so can't be split up. I'm assuming that there is no friction. Any thoughts would be much appreciated.

 

[Kevin Willis]

kinetic energy is a scalar, so can't be split up.

Kinetic energy, like speed, can be increased or decreased. When the ball hits the wall, part of its energy is used to heat the wall and the ball up, assuming it just bounces off the wall. The ball has lost energy in this interaction and given it to the particles in the wall (heat energy or molecular kinetic energy). Finding out exactly how much energy you need to know the angle, speed, composition of the ball and wall, and many more values. If you know speed before, speed after, and ball weight you can loosely determine the amount of energy lost.

 

[jbriggs444]

Since you are assuming no friction, it follows that there can be no component of the collision force that is parallel to the wall. The force must all be in the perpendicular direction. You are correct that this means that any motion of the ball parallel to the wall can be ignored in computing work done.

Although kinetic energy is a scalar, that does not mean that we cannot account for the total kinetic energy as partly due to motion in the x direction, partly due to motion in the y direction and partly due to motion in the z direction. The pythagorean theorem makes this work out nicely. We know that the square of total velocity is the sum of the squares of the velocity components in the x, y and z directions. So total energy is given by one half of mass times the sum of the squared velocity components in the x, y and z direction.

However, that's largely irrelevant to the question you posed. Recall the definition of work as the [vector dot] product of force applied multiplied by distance moved. If the wall is not moving that means that no work is done on the wall.

If the wall is moving then, in the absence of friction, the work done can be computed as distance moved by the wall (in the normal direction) multiplied by force applied by the ball (which is neccessarily in the normal direction).

We normally treat a collision as if it were instantaneous. So the force applied is infinite and the duration is zero. That means that the product of force applied times distance moved is technically undefined. However, it turns out that one can still solve such problems. The product of force applied times duration of collision is equal to the "impulse" applied to the ball. If you know how much the momentum of the ball changed then that is how much impulse was involved.

impulse(ball) = m(ball) * delta-v(ball)

By Newton's third law, the impulse applied to the wall is equal and opposite.

impulse(wall) = - m(ball) * delta-v(ball)

Recall the definition of impulse as force multiplied by time.

force-on-wall * collision-duration = - m(ball) * delta-v(ball).

Multiply both sides by the velocity of the wall

force-on-wall * collision-duration * velocity(wall) = -m (ball) * delta-v(ball) * velocity(wall)

But time times velocity gives a distance

force-on-wall * distance-moved-by-wall = - m(ball) * delta-v(ball) * velocity(wall)

And that's what we're after

work-done-on-wall = - m(ball) * delta-v(ball) * velocity(wall)

[There is some lack of rigor in the above, but it can be worked out rigorously and still gives the same result]

 

[sfensphan]

Thanks for the very detailed reply! I suspected that they would be separable (if that's the right word) for the case in which the wall is rigid but just didn't know what tool to use.

A followup if I could. What if the wall doesn't move but deforms? I don't think it would be as separable as before. Thanks again for your thoughts!