# Oxford PAT 2015, Question 18

## 9 thoughts on “Oxford PAT 2015, Question 18”

1. jason says:

why dont you simply use ft=m(v-u) and sub m/t = xp

2. TomYil says:

Hi, do you think you would get full marks if you write the expression for the force as f=mdv/dt and say m=pvAdt so the force for the first problem is pvA *-v = -pAV^2 and then substitute in v=x/A ?

1. Since the mark schemes are never published, I’m afraid it is unknowable whether you would get full marks for your answer. But having said that, your answer is basically correct reasoning, and you get the correct result — you can’t really do any more than that.

3. tara says:

Hi, I was wondering if it’s okay to answer a question like this by just manipulating the quantities to get the desired units. I did this and got the same answer, do you think I would be awarded method marks?

1. I don’t have any idea what the marking scheme is I’m afraid. But your solution sounds quite smart and getting the right answer in an ingenious way has got to get you a reasonable number of marks.

4. Tom Parker says:

Hi there, what you have done here makes perfect sense, but I approached this problem using energies rather than momentum, and my answers somehow gained a factor of 1/2. For i) if we assume the power of the water hitting the wall is equal to the kinetic energy transferred to the wall per second, and that the power is equal to the force of the water * the velocity, than surely the Force = (density*x^2)/2A. It’s probably obvious what I’ve done wrong but so far I’m stumped.

1. Tomasz Szawełło says:

In my opinion you should use average velocity. Then the power is equal to the force of the water * the velocity/2 and F=(density*x^2)/A 😉

1. Thanks for that suggestion: it does get you to the right answer, and it does also point to the problem in Tom’s answer above. It’s true that Power equals Force times Velocity, but what is the velocity of the water when it hits the wall? Just before, it is x/A and just after it is 0.

I think it is a bit dodgy just say “let’s just take the average of the two”, although it does work here. I would recommend that you use Force equals rate of change of Momentum with respect to Time. Remember, both Power = Fv and F = ma are really just special cases of F = d(Momentum)/dt, and when both m and_v_ are changing you are best off with the general case.