# Catching or deflecting an incoming ball for faster velocity?

1. Oct 15, 2007

### hawainpanda

1. The problem statement, all variables and given/known data
This is a conceptual question:
A ball is thrown to a person (at rest) on a skateboard. Should the person catch or deflect the ball to achieve the greatest velocity.

2. Relevant equations

3. The attempt at a solution
I thought it would be catching the ball, catching the ball means the force acting on the person for a longer duration of time and therefore larger impulse and larger change in momementum which means larger final velocity.

Thanks!

2. Oct 15, 2007

### BishopUser

probably better to think of this in conservation of momentum

$m_{s}v_{fs}+m_{b}v_{fb} = m_{s}v_{is}+m_{b}v_{ib}$

where s = skateboard, b = ball, i = initial, f = final

In each case try to imagine what would happen to the formula

3. Oct 15, 2007

### hawainpanda

I already did, using the conservation of momentum I'm right, but my instructor tells me other wise...funny thing is that http://www.sfu.ca/phys/120/PracticeMidterm2.pdf this says I'm correct, question 10.

4. Oct 15, 2007

### bob1182006

in that practice midterm you want to minimize the velocity and catching the ball does that. so you DONT want to catch the ball.

when you deflect the ball you exert a force on it to push it away, and it pushes you by Newton's 3rd Law so you speed up faster than if you just took the entire momentum of the ball.

I think that's how the problem meant to be, maybe wrong wording tho ><

5. Oct 15, 2007

### BishopUser

I think that answer key is wrong...
edit: ok the wording of the practice and your problem is opposite. on the practice exam you want to go slower and on your problem you want to go faster...

Case 1:
$(m_b+m_s)v_f=m_bv_{ib}$

Case 2:
$m_b(-v_{fb})+m_sv_{sf} = m_bv_{ib}$
$m_sv_{fs}=m_b(v_{fb}+v_{ib})$

Last edited: Oct 15, 2007
6. Oct 15, 2007

### hawainpanda

you're right, the midterm is the opposite, but I still think I'm right

I did a proof with the conservation of momentum and tried different values and they always proved that catching would create a larger delta momentum

7. Oct 15, 2007

### BishopUser

probably not plugging in a negative velocity when it shoots back the opposite direction

8. Oct 17, 2007

### tru_royalty

I'm quite ignorant of all the intricacies and complexities of this topic so, I'm not even sure if my thoughts have any merit here...I'm looking at the question's counterpart found at the link you provided...Some relatively specific statements are made. The ball is "heavy", it goes back to the source, and does so with the same speed at which it came..."Heavy" just hints at its large inertia, and large force required to deflect it (it's multiple choice). Mathematically: If ball is caught, mu=(M+m)v...and v=mu/M+m...right? If ball is deflected...MA=ma, and A=ma/M but a=(v-u)/t = -2u/t........................ So A= (-2um)/Mt...but A=(v-0)/t, so v= -2um/M...[Note that Uppercase letters are values for the boy, and lowewrcase for the ball.] Does this make sense? Does it not show that the value for deflection is greater?

Last edited: Oct 17, 2007
9. Oct 17, 2007

### HallsofIvy

Staff Emeritus
Once you've caught the ball, its momentum is no longer "acting on you". Your change in momentum is the opposite of the balls change in momentum (so that momentum is conserved). If the ball coming at you has momentum p and you catch it (so it now has momentum 0 relative to you) how much has its momentum changed? If the ball coming at you has momentum p and you deflect it back (so it now has momentum -p) how much has its momentum changed?

10. Oct 17, 2007

### tru_royalty

Of course, your question does not tell us that the ball has an equal but opposite velocity after reflection...But, I still believe deflection HAS THE POTENTIAL to give you a larger velocity...I don't know all the scientific fluff involved...but, logically speaking, the lack of details gives you an amazing amount of freedom. The fact is, it is POSSIBLE to deflect the ball with as great a force as you please (of even impractical, unlikely, and purely theoretical magnitudes). With deflection, the force with which the boy is propelled, and thus his acceleration is dependant on the force that he exerts, and lies within an almost infinite range. The acceleration he gains if he catches the ball is constant, while that which he COULD gain by deflection and its resultant is much greater. This may not be the ideal answer, but am I wrong?

Last edited: Oct 17, 2007