A 600 g air-track glider collides with a spring at one end of the track. FIGURE EX11.13 shows the glider's velocity and the force exerted on the glider by the spring. How long is the glider in contact with the spring?

Video transcript

Hey, everyone in this problem, a group of physics students are conducting an experiment to test the limits of sprints. They have a 400 g block moving on a frictional surface which collides with the sprint and the velocity of the block and the force exerted on it are recorded and shown in the figure that were given here. And we're asked to find the duration of the collision and the time that that block spends in contact with the spring. So we have two figures given here. The first one, we have the time in seconds on the X axis, the velocity in meters per second on the Y axis. OK. This is a purple curve. The velocity kind of goes from negative 2 m per second at the beginning. OK. It reaches this point where it's almost vertical before increasing up to m per second. OK. Kind of like this s shape. And then we have the second curve in green, we have time and seconds on the X axis, the force in newtons on the y axis and we have our force at zero until it increases kind of in this triangular shape up to newtons and then decreases back to zero. You know, over this time delta T that we are trying to find. So we have four answer choices in this problem all in seconds. Option A 0.36 option B 0. option C 0.13 and option D 0.22. So we're looking for this time and we have some information about velocity, we have some information about force. So let's think about our impulse momentum principles. OK. So we know that the change in momentum delta P is equal to the impulse J. You can also recall that the impulse J can be written as the net force multiplied by delta T. So this is gonna be delta P is equal to X multiplied by delta team. OK? And we're giving our force time curve, this is just gonna be equal to the area under that curve is our change of momentum. It's just going to be equal to the area under this curve in green that we are given. Now, let's kind of write out what the momentum is, what the area under this curve is. And see if we have everything we need to calculate this missing time that we're looking for. So the change of momentum recall is gonna be the final momentum minus the initial momentum momentum is given by the mass multiplied by the velocity. So we have MV two minus MV one. OK. So the momentum at this time 0.2 finds the momentum at time 0. this is gonna be equal to the area under the curve. And like I mentioned, when I describe this graph, this area is gonna be the area of a triangle. And so that's gonna be given by one half multiplied by the base multiplied by the height. Now we know the velocity, OK. Both before and after this increase in time. So we're showing that in our graph with the velocity of this purple graph here, we know the base of our triangle is delta T. So that's what we're looking for. And we know the height of our triangle. And so we have everything we need here to calculate delta T substituting in our values see the mass we're given is 400 g. We can convert that to kilograms by dividing by 1000. So we have 0.4 kg multiplied by the speed after this impulse. And after the impulse, we're on the right hand side of our curve here and we have a speed of about 2 m per second. We're gonna subtract the mass again, 0.4 kg multiplied by the speed before this impulse. OK. And that's gonna be negative 2 m per second. All of this is gonna be equal to one half multiplied by the base of our triangle. Delta T multiplied by the height of our triangle 25 noons. So simplifying on the left hand side, we're gonna have 1.6 kilogram meters per second. On the right hand side, we have one half multiplied by 25 newtons. So that's gonna give us 12 oops 12.5. There we go. And we have newtons. And we're gonna write that as kilogram meter per second squared. You remember that those units are equivalent and this is multiplied by delta T that time period that we're looking for. We're almost there. Last thing to do is to divide by this 12.5 kg meter per second squared. And we get delta T is equal to the kilogram meter per second is gonna divide it. We're gonna be left at the unit of seconds, which is what we want for time. And we get that delta T is about 0. seconds. So we look at our answer choices. We're going round to two decimal places and we can see that the duration of the collision. Ok? The time that the block spent in contact with the spring is about 0. seconds corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.