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Suppose a child drives a bumper car head on into the side rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

a) $-800 \textrm{ kg m/s}$

b) $-1.2 \textrm{ m/s}$

Solution Video

OpenStax College Physics Solution, Chapter 8, Problem 11 (Problems & Exercises) (1:57)

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Video Transcript

This is College Physics Answers with Shaun Dychko. This bumper car initially going in the positive direction say, hits the side rail and the side rail exerts a force of magnitude four thousand Newtons on the car for a time of 0.2 seconds. Now the force exerted on the car is going to be in the opposite direction to the initial speed. So this is going to be a negative force. This force exerted by the side rail is the net force since there are no other horizontal forces to consider we're told that we can ignore friction. So change of momentum is going to be the net force multiplied by the time that it's acting. That's negative four thousand Newtons times 0.2 seconds which is negative 800 kilogram meters per second, is the impulse. In part B we're asked to find out what is the final velocity of this car after the collision given that it has a mass of two hundred kilograms and an initial velocity of 2.8 meters per second which we'll take to be positive. So the impulse is the final momentum minus the initial momentum and we can solve for the final momentum by adding initial momentum to both sides. We're doing this calculation because the final momentum contains the velocity factor in it. So we can substitute <i>m vF</i> in place of <i>PF</i> and then divide both sides by <i>m</i> to solve for the final velocity of the bumper car. So its final velocity will be the impulse it experiences plus the initial momentum that it has, mass times initial velocity, divided by its mass. So that's negative 800 kilogram meters sper econd that we calculated in part A, plus two hundred kilograms times 2.8 meters per second initial velocity, all divided by 200 kilograms, giving a final velocity of negative 1.2 meters per second. The negative sign means that the bumper car is going to bounce off the side rail and go the other direction after the collision with a speed of 1.2 meters per second.


Submitted by PlaywtPhysics on Fri, 06/12/2020 - 13:31

Hello Shaun,
Thank you for your helpful videos. I am studying from your website right now and this is my first comment.
Why did you assume that the force on the car by the side rail is in the negative direction? I assumed it is in the positive direction (because the question left it ambiguous) and got a completely different answer in part b.

Submitted by ShaunDychko on Mon, 06/15/2020 - 12:43

Hello PlaywtPhysics,
I'm really glad you're finding the videos helpful! Thanks for the comment. I assume you got a different answer for part (b)? It's fine to choose the coordinate system however you like. Choose left to be positive is fine. Regardless of the choice of coordinate system, it's very important to be consistent. Choosing left as positive would mean the initial velocity to the right is negative. Could you check for that in your work?
All the best,

In reply to by PlaywtPhysics

Submitted by PlaywtPhysics on Mon, 06/15/2020 - 18:46

You are right. I forgot to account for the direction of the initial velocity, which is negative according to the coordinate system I picked.
Is it typical for physics problems to give you the magnitude of things, and then it's up to you to take care of the signs that follow from your own diagrams?

In reply to by ShaunDychko

Submitted by ShaunDychko on Mon, 06/15/2020 - 20:27

Awesome, glad it's figured out. Yes, it's typical that the student is left to sort out the signs, whereas the question will usually use words to indicate direction, such as "left" or "right". For one, it's more readable with words, but additionally, a negative sign would be ambiguous unless the question explicitly establishes the coordinate system. Doing so for every question would be tedious. Now that you understand the issue, I think you're in good shape.
Best wishes with your studies,

In reply to by PlaywtPhysics