INT A 550 g cart is released from rest on a frictionless, 30° ramp, 120 cm from the bottom of the ramp. It rolls down, bounces off a rubber block at the bottom, and then rolls 80 cm back up the ramp. A high-speed video shows that the cart was in contact with the rubber block for 25 ms. What was the average force exerted on the cart by the block?

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Hey, everyone in this problem. A science teacher uses the following experiment to show her students the influence of elastic materials on collisions she released from rest, a 125 g cube on a smooth incline making an angle of 15 degrees. The cube slides 75 centimeters hits a super elastic material placed at the lowest part of the incline and then goes up centimeters along the incline. The contact time between the cube and the elastic material is measured to be milliseconds by an ultra sensitive motion detector. We're asked to find the mean force exerted by the cube on the elastic material. We're given four answer choices. Option A 0.7 newtons, option B 2.6 newtons, option C 3.3 noons and option D 5. noons gonna start by drawing out what we have. So we have this incline here and it makes a 15 degree angle with the horizontal. We have our cube that slides down that incline and at the very end of the incline, it's gonna hit some sort of elastic material. OK. It's gonna kind of bounce off that elastic material and move back up that incline we're gonna take to the right to be positive. Now, what we have is we have some sort of mean force or average force exerted over a very short amount of time. OK? We have this very small contact time. So we, we want to be thinking about is impulse and recall that our impulse and J is equal to the change in momentum delta P. Now, we also know that we can write our impulse J as the average force multiplied by delta T. On the left hand side, we have f average multiplied by delta T. OK. Where delta T is that contact time. And on the right hand side, the change in momentum, can you recall that momentum is the mass multiplied by velocity? And so our change in momentum is gonna be the mass multiplied by va minus the mass multiplied by V B. Now here va indicates the velocity rate after the impact and V B is going to be the velocity rate before the impact. OK. So before the cube and the elastic material are in contact, so we wanna find this mean for so we wanna find F average, we know delta T and we know that contact time, we know the mass of the Q. OK. We can write these things out. So delta T is equal to milliseconds and we're called that 75 milliseconds. We want to convert this to our standard unit. So going from milli millisecond to second, we're gonna use 10 to the exponent negative three OK. According to are standard units. And so we're gonna multiply it by 10 to the negative three OK seconds per millisecond. The unit of millisecond divides out and we're just gonna write this as 75 times 10 to the exponent negative three seconds. Now, for our mask, the mass of the cube we're told is 125 g. The same thing with this, we wanna convert to kilograms. So we're gonna multiply and we know that in every one kg there are 1000 g. So we multiply by one kg divided by 1000 g, the unit of gram divides out. And what we're doing is essentially dividing by 1000 to get our kilograms, which is gonna be 0.12 and five kilograms. So we have those two values. What we need in order to find f average is these velocity VA and B B. So let's go ahead and try to calculate those using our U AM or kinematic equations first. And then we're gonna come back to this impulse equation to find the average force that we're looking for. So let's start by finding the velocity V B before the impact. OK. Now this cube is gonna start from rest on the incline. So the initial velocity is actually zero m per second. And the final velocity of this cube when it slid down, that incline is gonna be V B. What we want to find, OK. When it makes it to the bottom of that incline just before it hits the elastic material, that's gonna be that final velocity, which we're calling V B, which is a speed again just before it hits the elastic material. So that's what we're looking for. Now, the acceleration we have to consider some acceleration due to gravity. OK? Because we're on an incline. Let me go to our diagram because we're on an incline. Gravity is going to be acting straight downwards, which means that it has some X component and some Y component. So you can imagine projecting this down two are inclined and the X component is gonna be related through the opposite side of that angle. And so we have to take the sign of the angle multiplied by that force of gravity G or the gravitational acceleration. And this is acting in the positive direction K, the X component of that gravity is pointing to the right, it's helping slide that block down the incline. So this is a positive acceleration. So we have sun of 15 degrees multiplied by the acceleration due to gravity 9.8 m per second squared. Now, we know how far this travels as well. We're told that the block traveled 75 centimeters before it hits that super elastic material. And so delta X is gonna be 75 centimeters. OK? We want to convert this to meters to go, go from centimeters to meters. We're gonna divide by 100 and we get that this is 0.75 m. You know, we aren't given any information about the time. T um and that's not what we're looking for. So that's OK. We have three known values V not A and delta X, we're looking for V F. So we can just choose the kinematic equation that has those four variables in it. And that's gonna be the following V F squared is equal to V not squared plus two A delta X. OK. V F. Well, that's the value V B that we're looking for. So we have V B squared is equal to the initial speed of zero. So the first term on the right hand side goes to zero, we get two multiplied by sine of degrees multiplied by 9.8 m per second squared, multiplied by 0.75 m. All right. So we can work this out on our calculator. We have meters per second multiplied by meters which is gonna give us meters squared per second squared. So we get V B squared is equal to 3.80 meters squared per second squared. And we take the square root and we get that the velocity of the cube right before the impact is 1. m per second. OK. So we found a V B, one of those velocities we were looking for. Now, we need to do the same with the velocity right after the collision. OK. So we wanna find a VA as well. So let's give ourselves some more room to do that. And now we're gonna find va now, in this case, when we're thinking about the block traveling back up the incline va is actually gonna be the initial speed. It's gonna be the speed that it starts with at the bottom of the incline right after the impact. OK. The final velocity is going to be zero m per second. OK. That cube is gonna come to rest, it's going to stop. It's gonna reach some maximum height on that incline. So the final speed is gonna be zero the acceleration, we have the exact same magnitude of acceleration. But in this case, our block is moving um upwards. So it's gonna act as a negative acceleration. Now we're gonna leave it as positive because we've said to the right as positive. So the acceleration is still technically acting in that positive direction, but it is gonna work to slow down that cube. Now we have delta X which is the distance that it the block travels back up that incline. We're told that that is 45 centimeters, but we have to be careful here because this is going now in the negative direction compared to what we've indicated in our diagram, we said to the right is positive, this is moving to the left. And so our displacement is actually going to be negative 45 centimeters. OK. And again, to convert to meters, we divide by 100 and we get negative 0.45 m. So we have the same four variables V not V fa and delta X that we want involved in our calculation. We're gonna use the same equation as we did to find V B. So we have V F squared is equal to V not squared plus two A delta X. So we have zero is equal to va squared plus two sign of degrees multiplied by 9.8 m per second squared, multiplied by negative 0.45 m moving va squared to the left hand side, we have va squared is equal to and then everything on this side is gonna become positive. OK? You can imagine that we moved this to the left hand side and then just swapped signs. So we get 2.2, 82, 784 again meter squared per second squared. And then when we take the square route, we get this speed immediately after the collision negative 1.5, 1089 m per second. OK. And the negative makes sense. It's moving the block is moving now up the incline in that negative direction. Oops and that's not squared anymore. That's after taking the square root. All right. So we have our velocities before and after the impact, we can move on to our impulse equation. So remember from above, we had that the average force of average multiplied by delta T is equal to M VA minus M V B. So now we have f average the contact time delta T we wrote above it was 75 times 10 to the negative three seconds. This is equal to the mass 0.125 kg multiplied by VA which we just found negative 1.51089 meters per second minus the mass again, multiplied by V B 1.905, I 1. 5055 m per second. All right. And if we work this out, what we're gonna do is to solve for f average, we need to divide both sides by 75 times 10 to the negative three seconds. Our units here we have kilogram meter per second, we're dividing by second. So we're gonna get kilogram meter per second squared and recall that that's equivalent to a Newton. And so we get F average when we simplify and divide is going to be equal to negative 5.769 newtons. So you'll notice we have a negative force here. We want to do is we want to just take the magnitude of that force. OK? Depending on which way we chose as positive, we could have got a different sign and So we wanna do is take the magnitude of the average force. And when we do that, we're gonna get that, that magnitude is approximately 5.8 newtons. If we compare this to our answer traces, this corresponds with answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Conservation of Energy

Impulse and Momentum