A 20 g ball of clay traveling east at 3.0 m/s collides with a 30 ...

Question

A 20 g ball of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay? Give your answer as an angle north of east.

Accepted Answer

Hi, everyone in this practice problem. We're being asked to calculate the resulting speed and direction of the combined 320 g puck where we'll have two different pucks with masses of 150 g and 170 g. And the first puck is moving to the east at 18 m per second. And the second puck is moving to the north at 24 m per seconds after the collision, the two pucks actually stick together And then move as one object. We're being asked to calculate the resulting speed after the collision and the direction of the combined 320 g. Puck expressing the answer as an angle north of east. The options given are a 15.3 m per second with an angle of 56.5 degrees north of East B, 12.1 m per second with an angle of 46.1 degrees north of east C 18.6 m per second with an angle of 38.5 degrees north of east. And lastly d 14.2 m per second with an angle of 65.6 degrees north of East. So we are given a bunch of different information. I'm going to differentiate it the two bucks with part one and part two. So first looking at the first pack M one will have the mass of 150 g or essentially 0. or 0.15 kg. And then the first buck is moving to the East at 18 m per second. So the velocity of the first buck is going to be m per second for moving to the east or in the eye factor notation direction. Next, the second book will have the mass of 170 g or .17 kg. With the velocity of the second puck being moving uh 24 24 m per second to the north. So that will be 24 J factor direction meters per seconds just like. So, so to find the final resulting speed and the direction of the combined Pike puck, we want to use the conservation of momentum. So the principle of conservation of momentum is that the total momentum of the first or the initial state will equals to the total momentum of the final state. So in this case, using the conservation of momentum, we will be able to find the resulting final speed after the collision or the combination of the two bucks. So the first or the initial will be P I equals to P F and initial momentum will be the momentum of uh the first part and the second buck separately. So that will be M one multiplied by V one plus M two multiplied by V two. You want to recall that momentum can be calculated by multiplying the mass with the velocity of our system, which is exactly what we are doing in this case. So the final momentum will be the momentum of the two buck combined, which is going to be M one plus M two multiplied by the final velocity that we're interested at, which is going to be this U right here. So we can actually pluck off uh uh pluck all of our known information into our equation right here. So M1 is going to be 0.15 kg multiplied by V one which is going to be 18 I meters per second Plus M2, which is 0.17 kg multiplied by 24 Jm for seconds. That will boil be all equals to 320g or essentially 0.32 kg multiply that by you, which is what we are interested at. So our you will then be um 0.15 kg multiplied by api meters for seconds Plus 0.17 kg multiplied by 24 J meters per second. All of that divided By 0.32 kg. So we want to divide 0.32 kg from the first term that we have and divide 0.32 kg from the second term that we have as well. And that will actually give us a value of U to be 8.4375, I Plus 12. J meters per second, just like so awesome. So we find our speed or our velocity right now in terms of the I and the J direction, but what we want or what we are interested at looking is actually just the magnitude of the speed itself. So we wanna find the magnitude of our philo U and that will be done by taking the square root of the first term or the I term squared plus the J term squared. So the velocity or the magnitude of the velocity or the speed is going to be the square root of 8. squared plus 12. squared. And that will be in meters per second. That will actually doing the calculation will actually give us a U value of 15.3 m per second. Awesome. So now what we are interested at looking for is the direction of The movement of the combined pack. So the di to calculate the direction we are going to utilize the right here, which can be calculated by taking the in first of tangent or tangent to the power of -1 Of the J Component Divided by the I component. So this will be 12.75 divided by 8.4375. And that inverse tangent value will actually correspond to a degree of 56.5 degrees north off east just like. So, so with the direction of 56.5 degrees north of east and the total or, or the actual speed of 15.3 m per second, the answer will actually correspond to option a where the resulting speed of the movement of the combined puck is going to be 15.3 m per second with a direction of 56.5 north of east. And that'll be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our art lesson videos on similar topic and that'll be all for this one. Thank you.

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A 20 g ball of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay? Give your answer as an angle north of east.

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