Ball hitting racket - Momentum Question (ENGAA 2017)

[TomK]

Homework Statement

ENGAA 2017 - Question 52

Relevant Equations

Force = rate of change of momentum

Please scroll-sown to Question 52: https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf

The correct answer is 'B'. This is the working I did:

F = (change in momentum) / (change in time)

change in momentum = mv - mu, where v = final velocity and u = initial velocity

initial velocity = +u, final velocity = -v (as velocity reverses direction after hitting racket)

change in time = t2 - t1m (-v - (+u)) = F (t2 - t1)
-mv - mu = Ft2 - Ft1
mv = Ft1 - Ft2 - mu
mv = F (t1 - t2) - mu

As you can see, my answer is wrong. I don't understand why. What's strange is that if you assume final velocity = +v and initial velocity = -u, you'll get the change in momentum to be m (v - (-u)) = mv + mu. Then, you'll get the right answer after rearranging.

 

[PeroK]

As you can see, my answer is wrong. I don't understand why. What's strange is that if you assume final velocity = +v and initial velocity = -u, you'll get the change in momentum to be m (v - (-u)) = mv + mu. Then, you'll get the right answer after rearranging.

What's strange about that?

 

[TomK]

What's strange about that?

Why does it work one way but not the other? The velocities are still in opposite directions. What error have I made?

 

[NTesla]

Try using vector. Then you will definitely get the right answer. Focus on the initial direction of the ball.

 

[PeroK]

Why does it work one way but not the other? The velocities are still in opposite directions. What error have I made?

First, let's analyse your answer. Note that you are asked for the magnitude of the final momentum. You have: $$mv = F(t_1 - t_2) - mu$$ From the question $u$ is a speed, hence positive, as is $F$ and $t_1 < t_2$. So, the expression you have on the right-hand side is negative. Which simply cannot be a magnitude. You should, therefore, have taken the modulus of this in any case to get a possible answer.

Second, let's analyse the question. If the initial speed $u = 0$, then the magnitude of the final momentum is: $$mv = F(t_2 - t_1)$$
Now, what happens when $u > 0$? Is the final momentum greater or smaller? The force must first stop the ball and then accelerate it, so some of the impulse is used to get the ball to rest, leaving less impulse to accelerate it. That means we have to take away the $mu$, leaving: $$mv = F(t_2 - t_1) - mu$$ Note that we have used the assumption that $F(t_2 - t_1) > mu$.

Finally, as suggested above, you could do a vector calculation. In which case, you need to use a vector for force. And, depending on the direction you take as positive, you will have $\vec F = \pm F$. And that's your mistake. You take the force to be positive, regardless of the direction you consider as positive for your velocities. In one case you get the right answer, but when $F$ has the wrong sign you get the wrong answer.

 

[TomK]

First, let's analyse your answer. Note that you are asked for the magnitude of the final momentum. You have: $$mv = F(t_1 - t_2) - mu$$ From the question $u$ is a speed, hence positive, as is $F$ and $t_1 < t_2$. So, the expression you have on the right-hand side is negative. Which simply cannot be a magnitude. You should, therefore, have taken the modulus of this in any case to get a possible answer.

Second, let's analyse the question. If the initial speed $u = 0$, then the magnitude of the final momentum is: $$mv = F(t_2 - t_1)$$
Now, what happens when $u > 0$? Is the final momentum greater or smaller? The force must first stop the ball and then accelerate it, so some of the impulse is used to get the ball to rest, leaving less impulse to accelerate it. That means we have to take away the $mu$, leaving: $$mv = F(t_2 - t_1) - mu$$ Note that we have used the assumption that $F(t_2 - t_1) > mu$.

Finally, as suggested above, you could do a vector calculation. In which case, you need to use a vector for force. And, depending on the direction you take as positive, you will have $\vec F = \pm F$. And that's your mistake. You take the force to be positive, regardless of the direction you consider as positive for your velocities. In one case you get the right answer, but when $F$ has the wrong sign you get the wrong answer.

Thank you. I now know I should have given F a '-' sign in-front, since force is acting in the negative direction, as I assumed positive to be right to left.

initial velocity = +u
final velocity = -v

F = change in p/change in t

-F = (-mv - mu)/(t2 - t1)
-F(t2 - t1) = -mv - mu

mv = F(t2 - t1) - mu
This is the correct answer.