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Problem 70 Easy Difficulty

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 200 $\mathrm{kg}$ and is traveling east with a velocity of magnitude 5.00 $\mathrm{m} / \mathrm{s} .$ Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. (a) A $25.0-\mathrm{kg}$ mass is thrown sideways out of the car with a velocity of magnitude 2.00 $\mathrm{m} / \mathrm{s}$ relative to the car's initial velocity. $(b) A 25.0-k g$ mass is thrown backward out of the car with a velocity of 5.00 $\mathrm{m} / \mathrm{s}$ relative to the initial motion of the car. (c) A 25.0 $\mathrm{kg}$ mass is thrown into the car with a velocity of 6.00 $\mathrm{m} / \mathrm{s}$ relative to the ground and opposite in direction to the initial velocity of the car.

Answer

(a) $v_{2}=5 \mathrm{m} / \mathrm{s}$
(b) $v_{2}=5.714 \mathrm{m} / \mathrm{s}$
(c) $v=3.78 \mathrm{m} / \mathrm{s}$

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Video Transcript

{'transcript': "Okay, so we're dealing with a railroad hand car moving along a track that is straightened frictionless. We're going to analyze three different situations here. So situation A is a 25 kg mass being thrown directly sideways out of the car. We have a 25 kg mass being thrown backwards out of the car, and we have a 25 mg mass being thrown into the car simulating basically a head on collision. So with situation A, we have no portion or no component of that moment. Um, from the from the peace being thrown out of the car, in the descent direction as or opposing the direction of the railroad car. So we have change it all. So the final speed remains five for, uh, scenario a now forcing area B. We're going to be dealing with basically an explosion. It acts a lot like a completely inelastic collision, but it's completely reversed. So our initial momentum that we need to deal with deal with here looks like this. So we're dealing with essentially initial momentum equaling final momentum where our initial is a massive 200. When they've lost E of five, our funnel mentum will be 1 75 moving at some unknown final speed plus a massive 25 moving at negative five meters per second. So now in solving this, we can go ahead, multiply out these two pieces so he would have 1000 on the left, equal to 1 75 Tom's of the final minus 1 25 and then added 1 25 to both sides. 1125 equals 175 times in the final. Our final speed here ends up being roughly 5.714 music per second for situation be and now moving on toe situation. See, we have a completely inelastic collision, a head on collision, so it will be the same idea. I said it got being a collision here. So our initial momentum is going to be a massive 200 moving at a speed of five, plus a mass of 25. Moving at a speed of negative sits because the direction is the opposite direction. Then this will equal a combined massive 25 moving at some unknown final speed. So now the left hand side. If we multiply all this out we will end up getting 1000 minus six times 25 850 on the left hand side. This is able 2 to 25 times the final speed now, dividing that by 2 25 we get a final speed of 3.778 meters per second."}