Hitting tennis balls against a wall A 0.058-kg tennis ball, traveling at $25 mathrmm / mathrms,$ hits a wall, rebounds with the same speed in the opposite direction, and is hit again by another player, causing the ball to return to the wall at the same speed. The collision of the ball with the wall lasts $1 mathrmms$. The ball returns to the wall once every $0.60 mathrms$. (a) Determine the average force exerted on the ball during a single collision and the force that the ball exerts on the wall averaged over the time between collisions. State the assumptions that you made. (b) If 10 people are practicing against a wall with an area of $30 mathrmm^2,$ what is the average pressure of the 10 tennis balls against the wall?
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So in this problem, we consider a tennis ball. Off Mass M is equal to 0.58 kilogram, which hits a wall with the speed B is equal to 25 meter per second and rebounds. Elastic Lee. The collision with the world lasts one millisecond, and the bone collector devolved once every 0.6 second. Let's do part in well tow obtain the average force exerted on on the ball. We use conservation of momentum. So be I. Less G is equal to final momentum. So initial momentum right be I is equal to M times. We 0.58 multiplied by 25 which is equal to 1.45 kilogram meter for a second. Well, the initial momentum points towards the wall and the final momentum points away from the world. The momentum A J is the momentum supplied by the wall and it can be written is J is equal to if one times delta D one here, if one is the average force over, uh, the course of the impact Delta T one, which is equal to tend to the power minus three seconds, is the time off the collegian. Then we obtain minus m times. We less if one times Delta T one is equal to 10 times we. And from here, if one is equal to twice off them, we divided by Delta T one Delta T one. Right now, let's like in numbers, so to multiply by 1.51 point 45 divided by 10 to the power minus three. So if one is equal to 2.9 kg Newton, right, well, the average force between the collisions is obtained in the same way way. Just use J is equal to J is equal to F two. Multiply by Delta T to Delta T two. Now, if two is equal to F two is equal to twice off em. We divided by Delta T two, which is equal to two, multiplied by 1.45 divided by 0.6, which is equal to 1.83 Newton, right, Mexico to 1.83 Uh, so it's 4.83 Newton, not one. It's four point in three Newton. Now let's do part. We've well to calculate the average pressure on the wall off surface Area A is equal to 30 m square with 10 balls colliding Village way embraced the pressure over the entire time between each collision. Because the collisions repeat every 0.6 seconds, we again use conservation of momentum tow. Obtain the average force for 10 balls between each collision. We is IAM old bold struggle at the same speed and our fire at the same time, and therefore we have minus 10 times. AM we less f times Delta Tito is equal to 10 times and we so f is equal to 20 times. And we divided by Delta D two. Let's like in numbers 20 multiplied by 1.45 divided by divided by 0.6. Therefore, F is equal to 48 0.3 Milton, 48.3 Newton and a great pressure.
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Now the average pressure B is equal to F, divided by a bridge force divided by area 48.3, divided by a DEA 30 m square, which is equal to 1.61 classical. Right In this entire calculation, we have assumed that no energy goes into deforming the ball different the impact and dead the ball travels perfectly horizontally