If a train hits a ball at $200$mph how fast does it bounce off the train?

Question

So I'm taking some classes on Brilliant and I'm starting with the basic stuff I already know and learned in hs/ms. We come to this question where we're dealing with switching frames of reference.

Now if the train is stationary the answer is simple. The $200$ mph ball bounces off and flies away at the same speed $200$ mph. Now if the train is moving brilliant says the ball winds up moving at $400$ mph. But this doesn't make sense to me.

The explanation is that from the train's perspective the ball is moving at it at $200$ mph and it bounces away at $200$ mph. then the train is moving at $200$ mph so to move to a stationary reference frame you add another $200$ mph. but this means that by changing nothing other than fiddling with how we look at it we took a $200$ mph ball and made it a $400$ mph ball. Because again: if the ball hits the train at $200$mph it leaves at $200$mph relative to the stationary viewpoint and the train.

Answer 1

but this means that by changing nothing other than fiddling with how we look at it we took a 200 mph ball and made it a 400 mph ball.

This is the core of what a frame of reference means. When talking about normal objects, their speed is not absolute. The ball doesn't have "a" speed. It has a speed relative to some other object or reference frame.

If you're in the cabin of a train and throw a candy bar at your companion facing you, you might estimate that while in the air, the bar has a speed of 5 mph. But someone watching the train speed past might say the bar has a speed of 50 mph. The speed "changes" by nothing other than how we look at it.

You have correctly stated that because the train is so large, the collision doesn't change its speed much. So before and after, the ball will appear to have about the same speed relative to the train, but in opposite directions.

To find the speed in any other reference frame, you just need to add the relative velocity.

i still don't understand why its faster than that if the train is the thing that is moving?

Ah. There is only one frame where the speed in equals the speed out for an elastic collision, and that is the frame where the center of mass of the system is at rest.

For the system of train + ball, the mass difference is so huge that we just ignore the ball entirely. The COM of the train + ball is almost exactly the same as the COM of just the train. But if the two objects were closer in mass, you'd want to find the frame where the center of mass is at rest.

• So take your current speeds and translate them into the frame where the center of mass is at rest.
• In that frame you have an elastic collision, so both objects reverse their direction and keep their speed.
• Then you translate back into the frame where you want the answer.

Doing this for the example where we start in the ground frame with the ball velocity of zero and the train velocity of 200mph :

Frame	$\Delta v$ to ground	Train	Ball
Ground	0	+200mph	0
COM (before)	+200mph	0mph	-200mph
COM (after)	+200mph	0mph	+200mph
Ground	0	+200mph	+400mph

We find the frame where the COM is at rest. We subtract that velocity from both parts to find their velocity in that frame.

For the collision, we reverse their direction (change the sign for velocity components). The train velocity is zero before, so is zero after. The sign of the ball's velocity changes.

Then to translate back, we add back the frame difference to both objects velocities.

Answer 2

There is another way of explaining it using the equation for elastic collision. If two objects of mass $m_1$ and $m_2$ with velocities of $u_1$ and $u_2$ respectively collide elastically, their resulting velocities $v_1$ and $v_2$ are $$v_1 = \displaystyle\frac{m_1-m_2}{m_1+m_2} u_1 + \frac{2m_2}{m_1+m_2} u_2 \\ v_2 = \displaystyle\frac{2m_1}{m_1+m_2} u_1 + \frac{m_2-m_1}{m_1+m_2} u_2$$ In this case, $m_1 \gg m_2$ and we have $$v_1 \approx u_1 \\ v_2 \approx -u_2 + 2u_1$$

So the velocity of the lighter object reverses direction and also increases by twice the velocity of the heavier object, which agrees with the reference frame method.

Answer 3

I'll address two questions:

1. What actually happens to the tennis ball in the tennis ball's frame of reference when the tennis ball starts off stationary and is hit by a train at 200 mph.

2. How to get a tennis thrown at a stationary train to 400 mph just by changing reference frames.

1. Stationary ball, moving train, ball's reference frame

Just before the train hits the ball, the ball is not moving. This means that, when they collide, the train exerts a force upon the ball to accelerate it. Putting a force on the ball causes it to compress and deform. As long as the ball is moving slower than the train, there is a force on the ball since the ball can't pass through the train. However, once the ball has reached the train's speed (200 mph), there is no longer any force on the ball from the train. The ball is now free to re-expand into its initial shape. Since the ball is compressed against the front of the train, the ball expands and pushes off the train, resulting the the ball being propelled off the front of the train to a faster speed. The amount of pushing can be difficult to calculate, so we use the frame-of-reference trick so we can calculate using the simpler collision with a stationary train to find the answer.

2. Moving ball, stationary train, ball's reference frame

The tennis ball is thrown at 200 mph at a stationary train. If we use the ball's frame of reference, the the ball will be stationary and the train oncoming at 200 mph. If we keep the same reference frame after the collision--in other words, we keep flying passed the stationary train at 200 mph--the ball will being going 400 mph in the other direction: 200 mph with respect to the train plus another 200 mph due to our reference frame.

Answer 4

In both cases the speed of the ball changes by 400mph. In the first case it changes from 200mph in one direction to 200mph in the other, which is a total change of 400mph. In the second case it changes from 0 to 400mph.

If you observed the experiment in a helicopter moving relative to both the train and the ball you would see a different set of speeds, but again the change in the speed of the ball will still be 400mph.

Answer 5

The rule here is no matter what the ball changes speeds by a total of 400 mph.

The only thing that's inconsistent is how fast the ground is moving relative to the two objects in question.

Answer 6

The ball would be moving the same speed as the train after being hit. And from the passenger's perspective (still in the same moving train at the same speed) the ball would just hover in front of the train. Not looking like it was speeding away in front.