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A $2.00-\mathrm{kg}$ stone is sliding to the right on a frictionless horizontal surface at 5.00 $\mathrm{m} / \mathrm{s}$ when it is suddenly struck by an object that exerts a large horizontal force on it for a short period of time. The graph in Fig. 8.34 shows the magnitude of this force as a function of time. (a) What impulse does this force exert on the stone? (b) Just after the force stops acting, find the magnitude and direction of the stone's velocity if the force acts (i) to the right or (ii) to the left.

(a) $J=2.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$(b) $(\mathrm{i}) \overrightarrow{\mathrm{v}}_{2}=6.25 \mathrm{m} / \mathrm{s}$ to the right(ii) $\overrightarrow{\mathrm{v}}_{2}=3.75 \mathrm{m} / \mathrm{s}$ to the right

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

Lectures

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problem. 8 13 We have a stone sliding along a frictionless surface. It's going five meters per second at the beginning of time as a massive two kilograms. And then this stone is moving along, minding its own business, and then some rather large force wax it for a millisecond. Um, so we need to figure out what what the aftermath of that is going to look like. So the first thing to note is the units of the scale of this. Does this contribute up easily If you don't pay close attention, like you should with any grab the time here is in milliseconds. So this is only acting over a millisecond. And this isn't killing a Newton's. So this is actually 2500 Newtons. Obviously, we need the right numbers to get the right numbers out of you know what we do here. So the first part that we need to do, he's just to find the impulse. And this has nothing to do with the Stone because, as we know, the impulse for a constant force is just equal to the value of that forest times the amount of time it's being. So we have 2500 Newtons. No one notices, and unsurprisingly, this comes out two and 1/2 Newton seconds also kilogram, meters per second, the same units of momentum which will become important soon. And then the direction of this impulse, because in general, impulses of vector quantity is going to be in the same direction as the force. That's a thing here that has a direction to it in space. But now we want to do is find out the magnitude and direction of the Stones velocity if this works is applied to the right or to the left. So we know that the impulse in addition to being equal to this expression of with the force, is also the change in momentum. And since the mass of the stone is changing, it's not getting hit hard enough to break it or anything. This is going to be M delta. And so Delta V. This doesn't depend on which direction the force is being applied is going to be the impulse provided by the Mass, and this is 2.5 seconds provided by T kilograms. And so the change in velocity is going to be one in 1/4 of meters per second. Now the direction of this change in velocity is the same as the direction of the impulse, which we said is the same as the direction of the force. So when the forces directed to the right stone is already moving, so we'll have plus Delta V and five plus one and 1/4 is six and 1/4 and similarly would have minus help. And this is 3.7. Hi. Now, both of these, this one. Obviously, since we're adding to the lawsuit, this is still going to be moving to the right, and since this isn't greater than or equal to the initial velocity, the stone will be slowed down, but it's not going to stop or reverse direction.

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