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Problem 106 Hard Difficulty

Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear (Fig. 56). Car A has a mass of 450 $\mathrm{kg}$ and car $\mathrm{B} 490 \mathrm{kg},$ owing to differences in passenger mass. If car A approaches at 4.50 $\mathrm{m} / \mathrm{s}$ and car $\mathrm{B}$ is moving at 3.70 $\mathrm{m} / \mathrm{s}$ , calculate $(a)$ their velocities after the collision, and (b) the change in momentum of each.

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 9

Linear Momentum

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Video Transcript

the fact that this is an elastic collision on DSO for an elastic collision. Um, you're the difference from your initial velocity, Zvi. A minus you. Beauty is equal to the negative of the difference between the final velocities. Ah, via prime CB front. So you have here that BB prime? Is it cool to be a minus? Beauty plus V a prime. And so this is Ah, Via is 4.5 aa minus baby 3.7 plus v a prime. So, Bebe is there 4.8 Plus he a prime meters per second. Therefore from momentum conservation, we have obviously a v A plus m b b initial Momenta is equaled. Final memento, eh? Maybe a prime plus m b B B prime and so Vehbi prime is 0.8 times v a prime. Therefore, in writing things down in terms of being a prime, we have v a prime, uh, Times M A plus m b is equal to m a v a plus m beauty times, point times Ah v B minus 0.8. Okay, um and so v a prime therefore, is ah m a v A plus and B times V b minus 0.8 over M A plus M b. Okay, and therefore V A prime is just, um is 4 50 kilograms, turns 4.5 meters per second. Uh, plus for 90 kilograms turns 2.9 meters per second. Uh, over 4 50 plus 4 90 uh, kilograms. So v a prime is 3.67 meters per second. Ah, therefore Avi be prime. It's this plus 0.8. So this is 0.8 plus 3.67 calls 4.47 meters per second. All right. And in part B, we're just calculating the change in momentum. Eso for air you have dealt appear is an eight times V a minus b a prime change in velocity stars must So this is 4 50 kilograms times 3.67 67 minus 4.5 meters per second, giving Delta P A of negative 3 75 kg meters per second. Ah, and similarly, for a Delta P B, you have m b times B B minus V b prime and so that's 4 90 kilograms times 4.47 minus, uh, 3.7 uh meters per second, Right Similar to here. Uh, and so this gives us to be 75 kilograms meters per second. Note that Delta Pew B is equal and opposite to Delta Pier, which means the momentum is conserved.

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Physics for Scientists and Engineers with Modern Physics

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Problem

In a physics lab, a cube slides down a frictionle…

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Physics for Scientists and Engineers with Modern Physics

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Zachary Mitchell

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Zachary Mitchell

Jared Enns

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Marshall Styczinski

Zachary Mitchell

Jared Enns

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics