Chapter 8 3. A bumper car with mass mA = 119.0 kg is moving...

Chapter 8 3. A bumper car with mass $m_A = 119.0$ kg is moving to the right with a speed of $v_{A1} = 4.4$ m/s. A second bumper car with mass $m_B = 83.0$ kg is moving to the left with a speed of $v_{B1} = 3.5$ m/s. The two cars have an elastic collision. Assume the surface is frictionless. a. Find the velocity (x components $v_{A2}$ and $v_{B2}$) of each car after the collision. b. Find the impulse (magnitude and direction) exerted on $m_A$. c. Is the impulse (magnitude) on $m_A$ greater, smaller, or equal than the impulse on $m_B$? d. The graph below shows the time dependence of the force during the collision. Estimate the average and the maximum force.

Transcript

00:01 This question here we have a car which is moving at 15 meters per second and it has a mass of 1 .53 times 10 to 3 kilograms and it hits a wall.

00:18 So i'll draw on the ball there and it rebounds going in the other direction at 2 .6 meters per second and we are asked to calculate the impulse delivered to the car during the collision.

00:32 So the equation for impulse, what impulse is force times time and it's equal to the change in momentum.

00:41 So change in momentum is mv, so the final momentum minus the initial momentum.

00:48 And what can do is take out m from both, so it's not going to me mv minus u.

00:53 So that's been 1 .53 times 10 to the 3 for the mass.

00:58 And then we have v sorry v being 2 .6 but shall just make this a bit clearer it's 2 .6 but just bear in mind that velocity is a vector therefore direction matters so we need to choose which direction is positive which direction is negative i'm going to say that the initial direction from left to right is positive that makes this minus 2 .6 meters per second because it's then going from right to left.

01:30 So v is minus 2 .6 and then minus 15.

01:36 And that's going to get you to two significant figures, 2 .7 or minus 2 .7 rather, times 10 to the 4 newton seconds, or you can use kilogram metres per second.

01:50 So again, if left to right is positive, then it's moving initially with a velocity of 15 meters per second, but that means it rebounds with a velocity of minus 2 .6 meters per second.

02:01 The minus just tells you it's moving in the opposite direction to where it came from.

02:06 Then for b, we want to work out the force.

02:09 So the force is changing momentum over time, change in momentum over time.

02:15 And that is going to be minus 2 .7 times 10 to the 4 divided by 0 .15 because the collision occurs over 0 .15 seconds.

02:27 And that means there is a force of minus 180 ,000 newtons on the car.

02:34 Again, the minus sign just means that the wall exerts a force from right to left.

02:39 And that should make sense because the car rebounds from right to left.

02:43 Then for part c, we are told, if the car comes to a stop after the collision during the same interval, find the average force exerts on the car.

02:52 So again, force is changing momentum over time.

02:55 But now the car is not rebounding, it's just stopping...

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