# Which ball exerts a larger impulse on the wall?

1. Oct 2, 2016

### Steven Byers

• Originally posted in a technical forum section, so missing the HW template
A 10 gram rubber ball and a 10 gram iron ball are thrown at a wall with equal speeds. The rubber ball bounces back at almost the same speed with which it was thrown at the wall, while the iron ball bounces back at half the speed with which it was thrown at the wall. Which ball exerts a larger impulse on the wall?

A. Both balls exert the same impulse on the wall because they were thrown at the same speed.

B. Both balls exert the same impulse on the wall because they have the same mass.

C. The iron ball exerts a greater impulse on the wall because iron is tougher than rubber.

D. The rubber ball exerts a greater impulse on the wall because it bounces back with a higher speed than the iron ball.

I think the answer is B because Impulse equals Force times Time, with Force equaling Mass times Acceleration. If both balls have the same mass, then shouldn't both balls exert the same impulse on the wall? Forgive me for being ignorant on the subject. I wasn't taught it yet and I'm making an educated guess based on my readings.

2. Oct 2, 2016

### Simon Bridge

Welcome to PF;
Force is rate of change of momentum ... but the question does not ask about force.
It asks about "impulse" ... impulse is just the change in momentum.

3. Oct 2, 2016

### Steven Byers

Hi Simon! Thank you for that clarification. If Force is the rate of change of momentum and momentum equals mass times velocity, then would the answer be D, considering the rubber ball bounces back with a higher velocity than the iron ball?

4. Oct 2, 2016

### Simon Bridge

You want to reason like this: the question is asking about impulse, impulse is change in momentum - therefore the ball with the greater change in momentum exerts the greater impulse on the wall (and receives the greater impulse from the wall). p=mv and both malls have the same mass throughout and the same initial velocity, so the ball with the greater final velocity exerts the greater impulse.

You don't want to reason from force because the relationship between impulse and force is not straight-forward ... during a collision, the force on the ball/wall varies over time. But if you want to try, then you have to express the force as a function of time: F(t), then integrate to find the impulse ... so, for a collision that starts at t=0 and ends at t=T, the impulse is the area under the F vs t graph: $I=\int_0^TF(t)\;dt$
... a simple model for F is quadratic during the time period and zero outside, something like:
$F(t)=-4t(t=T)F_{peak}/T^2$
... but you only know about velocities - you you will need $F=m\frac{d}{dt}v(t)$ so $I=m\int_{v(0)}^{v(T)} dv$ ... which gets you, indirectly, to the change in momentum reasoning but by doing more maths.

5. Oct 2, 2016

### Steven Byers

I would think that the rubber ball would have a greater change in momentum considering its change in momentum would be zero (since its initial and final velocities are exactly the same) and the change in momentum for the iron ball would be negative (since its final velocity would be negative, considering the ball is slowing down). Since zero is greater than any negative number, the rubber ball would have to exert a greater impulse on the wall. Is this the correct way to approach this conceptual question?

6. Oct 2, 2016

### haruspex

Velocity is a vector. The initial and final velocities of the rubber ball are a long way from equal, though the speeds are similar.
But when it asks which gives the greater impulse, you should read that as asking for the greater magnitude of impulse (because it does not define a positive direction).
So, figure out the changes in momentum, taking signs into account, then compare the magnitudes of the changes.

7. Oct 2, 2016

### Simon Bridge

... that would be correct if the problem statement said the rubber ball's initial and final velocities are the same but that is not what it says. It says the rubber ball bounces off with almost the same speed ... "bounces off" can safely be taken to imply a reversal of direction, and "speed" is, by definition, the magnitude of velocity.

Change in momentum for the rubber ball is approx -2mv wrt the initial direction and +2mv wrt the final direction. The steel ball has -m(u+v) or +m(u+v) taking u<v as the final speed.
So you see both balls end up with the same sign for change in momentum.

It is possible to interpret a larger negative number being smaller, after all: -10 > -100.
... in which case, the answer would be "none of the above". As Haruspex points out, it is safe to take the interpretation "magnitude of the impulse": it like when someone asks which is the greater force, 10N or -10000N?

The matter becomes clear when the vector notation is used ... in the shorthand above, the sign of the number represents the direction of the vector, not it's size. The "size" of a vector is it's magnitude - draw two arrows on a sheet of paper pointing in different direction and make one of them bigger than the other one. See?

Last edited: Oct 2, 2016
8. Oct 2, 2016

### Staff: Mentor

[Mod note: I've changed the thread title to make it reflect the question being asked. It now reads: "Which ball exerts a larger impulse on the wall?" ]