A mine car (mass =440 kg ) rolls at a speed of 0.50 m / s on a horizontal track, as the drawing

step 1 We have a little mine car moving at a speed of 0 .5 meters per second. I guess I should put some

A mine car (mass =440 kg ) rolls at a speed of 0.50 m / s on...

A mine car (mass $=440 \mathrm{kg}$ ) rolls at a speed of 0.50 $\mathrm{m} / \mathrm{s}$ on a horizontal track, as the drawing shows. A 150 -kg chunk of coal has a speed of 0.80 $\mathrm{m} / \mathrm{s}$ when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

Key Concepts

Inelastic Collision

An inelastic collision is one in which the colliding bodies stick together after colliding, meaning they share a common final velocity. Although kinetic energy is not conserved in this type of collision, momentum conservation still applies, allowing the calculation of the final speed of the combined mass.

Conservation of Linear Momentum

In any isolated system where external forces are negligible, the total momentum remains constant before and after an event such as a collision. This principle is used to relate the momentum of different objects before they interact (such as a moving car and a falling piece of coal) to the momentum of the combined system after the interaction.

Transcript

00:01 We have a little mine car moving at a speed of 0 .5 meters per second.

00:06 I guess i should put some wheels on it.

00:08 And it has a mass of 440 kilograms.

00:14 And then a piece of coal lands in the mine car, but the piece of coal is coming down at an angle.

00:19 So the coal has an initial speed of 0 .8 meters per second.

00:28 This is 0 .50.

00:29 And its mass is 150 kilograms.

00:35 So the coal goes into the mine car.

00:37 The mine car is going to slow down a little bit because momentum is conserved.

00:41 And our question asks us to find the final speed of the system after the coal lands in the mine car.

00:49 So it makes sense to draw a momentum diagram.

00:52 If we just draw the momentum vector of the coal, the initial momentum of the coal, and we know that momentum is just mass times velocity so i can write it over here.

01:03 Momentum of the coal is going to be 150 kilograms times 0 .8 meters per second, which is 120 kilogram meters per second.

01:17 So that would be this vector here.

01:22 And it's coming down at an angle, which i've shown, but i haven't indicated the angle here.

01:26 The angle is 25 degrees, so that would also be the same angle.

01:32 Up here, 25 degrees.

01:35 The wagon, which the momentum vector is going to be horizontal because that's the way it's moving.

01:42 P initial of the wagon is going to be its mass times its initial velocity, 440 kilograms times 0 .5 meters per second, and that is 220 kilogram meters per second.

02:04 220 here.

02:07 So what's going to happen when they combine the sum of the momentum vector before is equal to the sum of momentum after for the total system.

02:16 The problem is, or the differences here, the floor is going to exert a force upward on the coal.

02:23 The normal force is going to increase.

02:25 So really all we're interested in because we don't have any sideways forces, they mention no friction or retiring forces...

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