A 60 g tennis ball with an initial speed of 32 m/s hits a wall...

Question

A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. FIGURE P11.40 shows the force of the wall on the ball during the collision. What is the value of F_max , the maximum value of the contact force during the collision?

Accepted Answer

Hey, everyone in this problem during a football match, a player attempts to score a goal but misses and the ball with a speed of 55 m per second hits the goalpost and rebounds back with the same speed. The force exerted by the goalpost on the ball is depicted in the graph below. So we have a graph and on the X axis, we have the time in milliseconds. And on the Y axis, we have the X component of this force. And we're asked to determine the maximum force exerted by the goalpost on the ball. If the mass of a football is 425 g, when we're told to consider the motion towards the right as positive, if we go ahead and look at our diagram, can we kind of have this trapezoid looking function? So we have the force starting at zero increasing linearly until three milliseconds to this F max, this maximum force that we're trying to calculate. It then remains horizontal from 3 to 6 milliseconds and then decreases linearly back down to zero at nine milliseconds. We're given four answer choices all in Newton's option. A 5.19 multiplied by 10 to the exponent three. Option B is the negative of a option C 7.79 multiplied by 10 to the exponent three. And option D is the negative of C. So we have this force applied over a short amount of time. So let's think about an impulse. OK. And recall that the impulse J is equal to the force F multiplied by the change in time delta T, right. This is our impulse equation. Now, in terms of our diagram because we have the force multiplied by delta T for our impulse. This is just gonna be the area under our force time graph. Hm. We also know that the impulse J can be written as the change in momentum delta P which is equal to the momentum two minus momentum one. Yeah. All right. So if we put this all together, what this tells us is that the momentum P two minus P one is going to be equal to the area under RFT. Perfect. OK. All right. So to find this area, we are gonna break this thing up into three. So we're gonna have this first section, that's a triangle, the second section, that's a rectangle and the third section, that's a triangle. OK. So we're gonna have or one part two and part three of our area. So when we go to our equation recall that the momentum is gonna be the mass multiplied by the velocity. So we get the mass multiplied by V two minus the mass multiplied by V one. And this is gonna equal to the sum of the three areas. A one plus A two plus A three, right? All right. Well, what are all of these areas? Ok. We're gonna leave the left hand side alone for now, MV two minus MV one, our area A one, this is a triangle. So it's just gonna be one half multiplied by B one H one. Then we have our area of our rectangle which is gonna be the length multiplied by the width. We're gonna call that L two and W-2 just so it's clear that we're talking about that second segment and then we have our area A three which again is a triangle, one half B three H three. OK. So this is just simple math, simple areas of these three separate shapes. OK? Let's start substituting in some values here. So we are told that the mass is 0.45 kg. OK? We're told it's oops 0.425. We're told that it's 425 g and we divide that by 1000 to convert to our standard unit of kilograms. OK? We need to be in kilograms. So we end up with the final unit of newtons, we're gonna multiply by our final velocity. Now, our initial velocity is 55 m per second. We're told that this rebounds with the same speed but it's now going in the negative direction. And so our final velocity is gonna be negative 55 meters per second. We're gonna subtract again that mass 0.42 5 kg multiplied by the initial velocity 55 m per second. And this is gonna be equal to, on the right hand side, we have one half the base of our red triangle while we're going from zero milliseconds to three milliseconds. So that's gonna be three milliseconds. We wanna convert to our standard unit of seconds. So we're gonna multiply by 10 to the exponent negative three to go from milliseconds into seconds. Then we're gonna multiply by the height. Now, the height of this triangle is just F max that maximum force that we're interested in. OK. So now we can see where that maximum force is coming into this equation. We're gonna be able to calculate what that is. OK. Now, we're adding the area of the blue rectangle, the area of the base or the length three milliseconds up to six milliseconds. So again, that's gonna be three milliseconds. And then we're multiplying by the height of that rectangle, which is again, F max, OK. So we have F max multiplied by three, multiplied by 10 to the exponent negative three seconds. Once again, F max, that value we're looking for. And finally, we have this green triangle, we're gonna have one half again, the base is going to be three multiplied by 10 to the exponent negative three seconds. And we are multiplying by that height of the max force as well. OK. Just written that under as we've run out a verb. All right. Now, this looks a bit messy, but all we've done is on the right hand side, calculate the area of a few simple ships. OK? On the left hand side, we have our change in moment. So let's go ahead and simplify here. On the left hand side. When we work this out, we're gonna get negative 46 0.75 kilogram meters per second. On the right hand side, we're gonna have 0.006 seconds multiplied by F max. So we can see the last step in order to isolate for F max is to divide by the 0.006 seconds. And we get that F max is equal to about negative 779.167 Newton. All right. So let's compare this to our answer choices. We wanna write this in scientific notation with three significant digits. And when we do that, we can see that the correct answer is going to correspond with option D negative 7.79 multiplied by 10 to the exponent three noons. Thanks everyone for watching. I hope this video helped see you in the next one.

A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. FIGURE P11.40 shows the force of the wall on the ball during the collision. What is the value of F_max , the maximum value of the contact force during the collision?

Key Concepts

Impulse and Momentum

Newton's Third Law of Motion

Force-Time Graph

Watch next

A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. FIGURE P11.40 shows the force of the wall on the ball during the collision. What is the value of Fmax , the maximum value of the contact force during the collision?

Key Concepts

Impulse and Momentum

Newton's Third Law of Motion

Force-Time Graph

Watch next

A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. FIGURE P11.40 shows the force of the wall on the ball during the collision. What is the value of F_max , the maximum value of the contact force during the collision?