INT One end of a massless, 30-cm-long spring with spring constant 15 N/m is attached to a 250 g stationary air-track glider; the other end is attached to the track. A 500 g glider hits and sticks to the 250 g glider, compressing the spring to a minimum length of 22 cm. What was the speed of the 500 g glider just before impact?

Question

Accepted Answer

Hey, everyone in this problem. A group of physics students is conducting an experiment using an air track. They have a 35 centimeter long massless spring with a spring constant of 10 noons per meter attached to the track at one end and an air track car of mass 320 g. At the other end, the air track car is initially at rest. The students are gonna launch a second air track car mass 600 g toward the first. They're gonna collide stick together and cause the spring to compress by a distance of 20 centimeters from its natural length. And we're asked to determine the velocity of the 600 g air track car right before the collision. We're given four answer choices all in meters per second. Option A 2.4 option B 0.32 option C 0.8 and option D one. Now we're gonna think about this kind of in two parts. OK. So we have the collision part of this problem and then we have the post collision part where that spring is going to be compressed. OK? So let's think about the conservation of momentum that we have in the collision. OK. So let's think about the collision itself and we know that the initial momentum peanut is going to be equal to the momentum immediately after the collision, which we're gonna denote by P A. OK. So the knot subscript is gonna be our initial A subscript is going to indicate immediately after that collision. OK. But before we get into the spring compression, now our initial momentum, we have two air track cars. So that's gonna be made up of those two cars. So we have peanut, one plus peanut too. And our momentum after the collision is going to be just made up of one thing we're gonna call it P A because those air track cars stick together. Ok. So it's gonna be the total momentum of them stuck together. Now recall that momentum is gonna be the mass multiplied by the velocity. And so our equation becomes M one VN one M two VN two is equal to N A multiplied by the. Now we're interested in finding, OK? Is V two, the velocity of the 600 g air track car immediately after the push who are immediately before the, we're looking for V not two. Now V not one we know is going to be equal to zero because this car is initially at rest. Ok? So that term goes to zero, we're left with M two V, not two is equal to the mass after the collision which is gonna be the total mass M one plus M two of the two cars when they stick together multiplied by now again, V not two is what we're really trying to fight. But we also have a va value that we don't know. So we're gonna switch over to after collision and we're gonna look at the conservation of mechanical energy there and we're gonna see those terms popping up again. So let's go ahead and isolate one for the other. And what we're actually gonna do is we're gonna solve for va in terms of V two. So that when we substitute it into our equation later, our equation will be in terms of this V not two value we're interested in and we could substitute it the other way, but it's just gonna require an extra calculation at the end. What we're gonna do is we're gonna write VA is equal to N too multiplied by V, not two do right by M one plus M two. We're gonna call this equation one. All right. So we've dealt with the collision in the conservation of momentum. Now, we're gonna look at post collision, conservation of energy. OK. So post collision, we have a conservation of mechanical energy, as I just mentioned. So we have that the kinetic energy immediately after the collision off the potential energy immediately after the collision. And this is gonna be UAs to indicate the spring potential energy. OK? We have no gravitational potential energy to worry about because these cars are on the track. OK. So we only have our spring potential energy. This is gonna equal the fine kinetic energy KF plus the final spring energy eu F All right. Now, the initial kinetic or sorry, the initial spring potential energy uax immediately after that collision. Well, this is before the spring has been compressed. So the spring is at its natural length hasn't been compressed, it's in its resting position. So that energy is just going to be easier. So what we get is we have one half ma va squared. On the left hand side, the right hand side, we can eliminate our final kinetic energy. Because when the cars or when that spring gets to its maximum compression, the cars are going to be stopped, their velocity is going to be zero momentarily as they change direction. OK. And so the kinetic energy, the final kinetic energy at that maximum spring compression is going to be zero. So that term goes to zero. And on the right hand side, we're left with just this energy due to the spring which is given by one half K XF squared. All right. Now, there's one half, we can divide it. We're left with ma multiplied by va squared is equal to KXF squared. No ma is the mass after the collision. OK. These cars stick together. We've done this before in our conservation of momentum, ma is gonna be the sum of the two masses M one plus M two. And now we can substitute in equation one for VA and we already know what VA is in terms of V not two. So let's substitute that in. So this is gonna be M two V, not two divided by M one or M two, all squared. That's gonna be equal to that spring constant K multiplied by XF squared. All right. Now, we have this M one plus M two term in the numerator, we have M one plus M two squared in the denominator. So if we simplify, we're gonna end up with M two squared. V, not two squared divided by M one plus M two is equal to KXF squared. Now, remember this VNO two value is what we're looking for. OK. So let's go ahead and isolate for VNO two before we substitute all our values in. If you wanna substitute the values we have in first, you can go ahead and do that as well. So simplifying we're gonna multiply both sides by M one plus M two. And then we're gonna divide by M two squared and we get that V not two squared, it is going to be equal to M one or M two multiplied IKXF script divided by M to sweat. OK. So V not to, the final value is going to be the square root. Everything we've just written M one was M two multiplied by KX squared. Divided by M two squared. All right. Now we're looking for a speed. So we're taking the positive route here and we don't care about the direction and we can kind of infer the direction as well from the problem. So we're gonna just look at the positive route here and now we can go ahead and substitute in all of the values. We know when doing that. What do we have? We have that V not two is going to be equal to the square root of 0.32 kg. We were given the car masses in grams, we wanna write them in kilograms, our standard unit. So to do that, we divide by 1000. OK. So we have 0.32 kg was 0.6 kg multiplied by the spring constant 10 newtons per meter multiplied by XF squared, which is that compression and we go back up to the problem. We can see that the spring compresses by 20 centimeters. So that's gonna be 20 centimeters converting into meters or standard unit. We divide by 100 we get 0.2 m squared and all of this is divided by the mass M two squared and that is at 600 g air track car, the one that's initially moving. So 0.6 kg all squared. Now, if we work all of this out on our calculator, we are gonna get the the initial speed of car two, that 600 g air torque car is approximately 1.01 m per second. Bye. All right. Now, going back up to our answer choices and they're all positive. So we're gonna choose that this initial direction that this car is moving is the positive direction. So we have a positive velocity. When we compare our answer choices, we can see that this corresponds with answer choice D about 1 m per second. Thanks everyone for watching. I hope this video helped you in the next one.

Verified Solution

Key Concepts

Conservation of Momentum

Hooke's Law

Kinetic Energy and Potential Energy