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In a performance test, each of two cars takes 9.0 s to accelerate from rest to 27 m/s. Car A has a mass of 1400 kg, and car B has a mass of 1900 kg. Find the net average force that acts on each car during the test.

5700 $\mathrm{N}$

Physics 101 Mechanics

Chapter 7

Impulse and Momentum

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Winnipeg

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All right. So question number four, we're comparing the net force that is acting on each car during a test. So this performance tests we have two cars, they're both accelerating to the same final velocity. However, there masses are different. So we're going to compare the force that must act on both of these cars to figure out the amount of force needed to get to that final. Next, we'll see. Okay, let's start with recording information that we know. So in this problem we have two cars. Call the massive car. Eh is 1400 kilograms. The mass of Carby is 1900 kilograms. It takes them a total time of nine seconds. That would be our change in time here that the forces acting over we know that we're starting at rest from the problem. So I'll be initial is going to be zero meters per second. Like I mentioned earlier. Both cars will have the same final velocity. So are the final is equal to 27 meters per second and who were trying to find is the force that is acting on both cars. One challenge six this year. There we go. So you're trying to find is the force car, eh? And the force on car B. Okay, so to do this, we're going to use our impulse momentum term. So we know that our change in momentum is equal to the average force exerted over a period of time. It will be using this equation for both cars. So we're gonna work through one of them, and then we'll just change the numbers into the 2nd 1 So let's start with Carrey first. A couple things We know what we know. That Delta PR change momentum is the same as P final minus P initial soapy final. Nice p initial will equal F Delta T and we know that we're starting from rest. So that means this p initial is zero. Momentum is already over here. Momentum equals mass times velocity. So, for RP final, this is going to be the mass. And we said we'll start with car, eh? Massive car, eh? Times of lost. The final equals the force exerted times Delta t. Okay. What we're looking for is the force that's working. Divide delta t by both sides. So mass car, eh? Times final velocity all divided by Delta t equals the amount of force that was applied. So it's substitutes in numbers. In the massive car A is 1400 kilograms. Okay, times the final velocity, which is 27 meters per second, all divided by the change in time, which was nine seconds. So the force that's acting on car, eh? Label this one, eh? For sure is 4000 200 Newton's Now we do the same thing for Carby. Over here. The only thing that changes is the mass. So the force carby, we're gonna substitute in a different mass, which is 1900 kilograms times the final. Lawson, You remember they were both ending up with the same final speed 27 meters per second and it took the same amount of time nine seconds. So the force that must have been exerted on Carby or by car B would be 5000 700 Newtons. So here we have the force on Car B. And here that's tagging on the end like this. We have car, eh? Area

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