In which balls are thrown, calculations are made, and further explanation is required.

Last time on The Physics Philes, we started discussing momentum and impulse. This week we’ll do an example problem, but before we begin I need to add a bit of information. As I was preparing for this week’s post, I realized that I didn’t provide the definition of impulse, which is kinda necessary to work out the problem. I skipped it last week because I don’t yet understand derivatives (although adan does give a wonderful explanation in the comments last week). But I’ll do my best.

It’s really nothing new. We talked about all the equations involved last week, I just didn’t put them together. Basically, we just have to notice where equalities exist. Do you remember these:

We can use all of these to figure out the impulse. The difference in momentum is equal to the impulse. It’s also equal to the net force times the difference in time. Therefore, impulse is equal to the net force times the difference in time. Also, momentum equals mass times velocity, therefore we can replace momentum with mass times velocity in the previous equations. In the end, we can use the component form of these equations to figure out impulse problems.

This suddenly makes perfect sense to me. I just hope I’m explaining it well. Maybe it will be more clear as we sink our teeth into this sample question.

Let’s say I throw a ball with a mass of 0.40 kg against a wall. It hits the wall going at 30 m/s and rebounds going 20 m/s. What is the impulse of the net force on the ball during the collision with the wall?

First thing we need to do is draw a diagram. Luckily, I’ve done it for you.

Notice how the before velocity is negative. That’s because it’s moving in the negative direction. (Don’t worry if you didn’t notice that initially. Neither did I.) We don’t need a y-axis because the only force is horizontal. We need to find the x-component of the impulse, so we’ll use the x-component equation. Since we know the mass and the velocity, we can find the momentum, and once we find the momentum we can find the impulse. Let’s do the before calculations first.

The mass of the ball is m = 0.40 kg. The velocity of the ball as it hits the wall is v = -30 m/s. So our equation for p1x is:

To find the after momentum, we just replace -30 m/s with +20 m/s:

Now that we know what the momentum for before and after, we can take the difference to find the impulse:

So the impulse of the ball during its collision with the wall is 20 N • s. Now let’s see if we can find the average horizontal force the wall exerts if contact with the ball lasts for 0.010 s.

We know from the definition of impulse that the impulse equals the net force times the difference in time. We’ve just figured out the impulse and we know that Δt = 0.010 s. So it’s a simple matter of plugging in the values:

So the average force the wall exerts on the ball is 2000 N. The force has to be that big so the ball’s momentum can change very quickly. I don’t know about you, but the concepts of momentum and impulse make a lot more sense after you see how it works in meatspace.

That’s about all I have time for today. Tune in next week, when I’ll explain (or mangle) conservation of momentum. That’s the plan, anyway. However, next week I start my ACTUAL, FOR REAL MATH CLASS WITH A GRADE AND EVERYTHING. Wish me luck.

Featured image credit: Schlüsselbein2007