Conservation of momentum and energy example

• [Instructor] Blocks A and B are pressed together with a spring between them. When the blocks are released from rest, the spring pushes the blocks apart so that the 0.75 kilogram block A moves up the 30 degree ramp to the left and the 0.25 kilogram block B moves to the right at a velocity of three meters per second. Assume friction is negligible.

Which of the following is the best estimate of the maximum distance d that block A slides up the ramp? So pause this video and see if you can figure that out.

All right, now let's just first think about what would happen intuitively. So we have A and B, could imagine you're squeezing them together so you have that energy stored in that spring. And then when you let go, we know that B moves to the right and you would intuitively think that A would move to the left.

And it seems like the law of conservation of momentum is going to be at play here. Because if the system of A and B is initially at rest, each of these blocks' velocities are zero; it doesn't matter what their masses are. That means that the total momentum for the system of A and B is going to be zero.

So even after we let go and B moves to the right and A moves to the left, the total momentum of this system should still be zero. And so maybe we can use that to figure out what A's velocity is as it's sliding to the left horizontally.

And then using that velocity, well we could think about what's its kinetic energy. And if we know its kinetic energy, we could think about, well, as it goes up the ramp, that kinetic energy's going to be turned into potential energy. And so how high up the ramp would it go? And then from that and that angle we could figure out our distance.

So let's do that. So let's start off with the law of conservation of momentum. We know that after releasing, A's momentum plus B's momentum should add up to zero because the sum of their momentums was zero when they were at rest together, when they were compressed.

So what we could say is the momentum for A is going to be A's mass. So let me write it right over here; so 0.75 kilograms is A's mass times its velocity. Well we don't know what this is. Let's call this velocity sub or final velocity for A, and that's going to be in meters per second, plus B's mass, which is 0.25 kilograms, times B's velocity. Well that's going to be three meters per second, times three meters per second. This should be equal to zero.

And so now we just have to solve for that. And so what do we get? Let's see, if we subtract this part from both sides, we will get, and this is just 0.75 kilogram meters per second, so we'll get 0.75 kilograms times A's velocity after it gets released is equal to negative 0.75 kilogram meters per second.

And so if you divide both sides by 0.75 kilograms, on the left, you're just going to be left with the velocity of A, and on the right, you would be left with negative one; then the kilograms would go away, negative one meters per second.

So A's velocity as it lets go from the spring or as it loses contact with that spring is going to be negative one meters per second. Now, what does the negative mean? Well, if to the right is positive three meters per second, then to the left is going to be negative. So it's going to be moving to the left at one meter per second.

So let's use that information to think about how far up the ramp it's going to go, and we're assuming that friction is negligible and nothing really interesting happens here, although we know in the real world it might bump a little bit, but we're just going to assume that that's not going to be a major factor.

And so all of that kinetic energy when it's moving to the left at one meter per second is going to be converted to potential energy at some height. So let's say that once all of the kinetic energy turns into potential energy, A is at an altitude of h.

So what's its potential energy going to be at that point in time? Well, it's going to be the mass of A times the gravitational field, roughly 9.8 meters per second squared; if we want to be even rougher, say 10 meters per second squared, times that height.

So that's going to be once all of our kinetic energy has been turned into potential energy. Well, this should be equal to the amount of kinetic energy we had at this point right over here. Well, what is that going to be? Well, that's going to be 1/2 m v squared.

So it's 1/2 times the mass of A times the velocity of A squared. So the velocity is negative one meters per second, and then we're going to square it. Well, we could divide both sides by the mass of A. Negative one squared is just going to be a one.

So this is just going to be a one eventually. And so you're left with; and then we can divide both sides by g, and we see that our height is equal to 1/2 and then we're going to have, divided by g, divided by g, and then this is going to be meters.

And so what is this going to roughly be? Well, h is going to be approximately, let's say that g is approximately 10 meters per second, because they talk about estimating here. So this would be one over two times 10, and then we're gonna get meters.

So this is gonna be 1/20 of a meter. So h is approximately 0.05 meters. So you might be tempted to say, oh, look, there's an answer there, but remember they're not asking us this h. They're asking us this d.

And so here we just have to use a little bit of trigonometry. Where we say, okay, what trig ratio deals with the opposite of an angle and the hypotenuse? This is a right triangle right over here. Well, that would be sine.

So sine of 30 degrees, sine of 30 degrees, would be equal to the opposite side, which is h over d. We know that h is 0.05. Sine of 30 degrees, if you don't have a calculator, this is a good one to know, is 1/2, so you have 1/2 is equal to 0.05 divided by d.

We can multiply both sides by d, and then you'll have d over two is equal to 0.05, or multiply both sides by two, and you get d is equal to 0.10, or I could say roughly equal to since we are estimating. We had an estimate for g there.

And then this is choice D right over there.