# Oxford PAT 2013, Question 18

## 11 thoughts on “Oxford PAT 2013, Question 18”

1. Jonathan Martin says:

Hi! I’m looking at this question after Q17 in 2012, where I had gotten the answer wrong as I tried to reason that the mass reached the top of the runway and then immediately fell due to conservation of energy. I had assumed that energy conservation arguments were wrong for circular motion questions like this one, but how come you are able to use it in this case?
Thank you in advance for your help, and for this website – it really has been invaluable!

1. Looking back at 2012, Q17, it isn’t that ‘energy conservation arguments are wrong’ — energy is still conserved of course, but if you consider what is happening when the mass is passing some distance under the point T you can see that it is actually moving from right to left: the KE due to this motion and the PE due to its height add up to the same value as the KE that it had when it was stationary on the slope. If you look down the comments on that question, there are some pretty good conservation of energy arguments put forward.

Considering this question, funnily enough this is one where you need to be very careful about using conservation of energy. Here, there are two steps:
1. When the bullet hits the mass you need to reason using conservation of momentum to calculate the mass of the combined body, because some energy is dissipated when the bullet sticks in to the mass.
2. After the bullet hits the mass you can use conservation of energy to work out the change in height.

2. Paul Kottering says:

why cant you work out the KE of the projectile before entering the mass, and equate that to the GPE of the mass and projectile after in order to work out the height? Thanks.

1. You can do that. You should presumably get the same answer, yes?

Oops! Sorry that comment above was wrong — I’ve had too many comments to keep up with in the last few days! Actually you can’t reason by using conservation of energy here because the only way in which you can have a collision that absorbs the bullet is by dissipating energy in the collision.

This means that not all of the kinetic energy of the bullet is turned into kinetic energy of the (bullet plus mass) — some of it is dissipated as heat.

1. Paul Kottering says:

nope, you get 12.2m. Maybe energy is lost in deforming the mass etc? i think maybe we should assume the collision is inelastic, and use only conservation of momentum for the collision? I’m not sure.

1. Exactly right, see the other comment I made above. The collision is inelastic — there is no elastic collision that ends up with the masses stuck together. So kinetic energy is not conserved (because some of the initial KE is lost to heat).

3. dopplereffect eeeeeeeeoooowwwwww says:

Ek=0.5mvv=0.50.2122122=12.2122 Ek of projectile

Ep=Mgh=Ek M-combined mass

h=Ek/Mg=(12.2122)/(12.210)=122/10=12.2m

I can’t see faults in either solutions. Can you help?

1. See above — not all the kinetic energy of the bullet is transferred to kinetic energy of the (bullet plus mass), so the reasoning isn’t valid.

1. dopplereffect eeeeeeeeoooowwwwww says:

Thank you.

4. Rickardo says:

is it wrong to talk about how the intial ke will be converted to gpe here because when the bullet hits the ball the intial k.e is converted to BOTH ke and gpe as it hits the ball?

1. I don’t understand why you think that when the projectile hits the ball, the gpe changes.

The question says “projectile was moving at the height of the centre of the ball, and after .. it stops inside the ball” (presumably at the same height). So if you consider the centre of mass of the (projectile + ball) system, the height of the centre of mass doesn’t change during the collision.

So if the height of the centre of mass doesn’t change during the collision, there is no change in gpe. That is, the change in gpe is zero as the projectile hits the ball.