# Oxford PAT 2013, Question 17

## 7 thoughts on “Oxford PAT 2013, Question 17”

1. jason says:

how do you know the object with mass m is going up the slop instead of going down. it wasnt specified in the question

1. Hi Jason, thanks for your question. This is an interesting and subtle point. If you want to write down some equations, you’ve got to choose a direction for the acceleration. But, as you point out, there’s no way of knowing which direction you ought to choose.

But the good news is: it doesn’t matter what direction you choose, because the acceleration can be negative. Choose the ‘wrong’ direction and you’ll just get a negative number for the acceleration; so the equations work whichever direction you choose.

Have a look at the equation that we get to on line 8. The acceleration is g(M-m sin α)/(M+m). So when m sin α > M the acceleration is negative, and the mass m goes down the slope. So when the mass m is going down, we just get a negative upwards acceleration.

2. Since a= (Mg-mgsina)/(M+m) and a must be 0 for the masses to be stationary

Mg-mgsina = 0

Therefore Mg = mgsina, so M = msina

Wouldn’t this and the fact that they must be stationary in the first place, and not moving at constant velocity, be necessary for the marks?

p.s thanks for the great resource that is this website!

3. kyle says:

what is the condition for teh mass to be stationary? 0 acceleration and thus at equillibrum?

1. Yes that’s right. I guess to be strictly correct the conditions for the mass to be stationary are that its acceleration is zero, and it was not already moving (because if it were it would be moving at a constant speed).