# Kinetic energy of a ball when falling (projectile motion)

1. Feb 4, 2012

### paperdoll

1. The problem statement, all variables and given/known data

A ball of mass 0.5 kg rolls off a table 4 metres above the ground with a speed of 4 ms^-1.

a) what is the kinetic energy of the ball when it strikes the ground.
b) what is the kinetic energy of the ball just before it strikes the ground
c) what is the speed of the ball as it strikes the ground

2. Relevant equations

Ek=1/2mv^2
V^2=u^2+2as
Ep=mgh

3. The attempt at a solution
okay so to get a) I did Ep=mgh so when the ball strikes the ground, all Ep should be turned into Ek right?

So 0.5*9.8*4=19.6 J

for b) I know the horizontal final velocity component is 4 ms^-1. The vertical final velocity should be found using v^2=u^2+2as.

v^2=2*-9.8*-4
v=8.85 ms^-1

using pythagoras throem I got the final v to be 9.7159 ms^-1.

Ek=9.715^2*0.5*0.5=23.6 J

What I'm not sure about is for answer b) I got 23.6 and for a) I got 19.6. Shouldn't a) be a larger value than b) because with a), the ball has striked the ground. I don't know where I have gone wrong

2. Feb 4, 2012

### BruceW

Your calculations are good, but I don't think you realise what you've calculated.

Your answer of 19.6J is the change in kinetic energy from when the ball was rolling along the table to when it was just about to hit the floor. And your answer of 23.6J is the KE of the ball just before it hit the floor.

So I think you have calculated b) correctly. To work out the answer to question a), you're going to need to make some kind of assumption about the motion of the ball exactly when it is touching the ground.

3. Feb 4, 2012

### paperdoll

Would the assumption be that v=0 when the ball hits the ground? So in that case, Ek is zero then?

I thought that when the ball hit the floor, all the Ep would be converted into Ek? A bit confused but I understand what you say about the assumption I think.

4. Feb 4, 2012

### BruceW

Just before the ball hits the floor, all the Ep will have been converted into Ek. And the ball on the table had 4J of Ek. And you calculated the change in KE is 19.6J. So adding these together, you get 23.6J (this is a good way to check your answer of part b).

The assumption v=0 when the ball hits the ground is one possible assumption. And this might be a good assumption if the ground was a very viscous fluid. But as you've said, Ek would be zero, and so question a) would have a suspiciously easy answer.

I think the assumption they want you to use is less strict. (And its a very common assumption in these types of questions). Think about the vertical speed of the ball before and after collision.

5. Feb 4, 2012

### paperdoll

I've been trying to think but I'm still not sure about the assumption. Would it have something to do with the conservation of momentum?

6. Feb 4, 2012

### BruceW

I guess you could use conservation of momentum to explain what happens to the horizontal speed. (Also making use of the fact that the force from the ground is purely vertical, assuming no friction).

For vertical speed, while the ball is touching the ground, we can say the ball squashes, and undergoes an extreme upwards acceleration. So really, while the ball is touching the ground, different parts of the ball will have different values for the vertical speed, and will go through a range of different values.

So the vertical speed while the ball is striking the ground is actually a vague concept, which could have several possible answers. So I suppose that the 'vertical speed as the ball strikes the ground' is chosen more as convention, rather than by assumption.

The vertical speed of the ball is negative before collision and positive after collision. So, you can probably guess what is the conventional vertical speed of the ball as it strikes the ground. (its a simple answer, no working required).