The stoplight had just changed and a 2000 kg Cadillac had entered the intersection, heading north at 3.0 m/s , when it was struck by a 1000 kg eastbound Volkswagen. The cars stuck together and slid to a halt, leaving skid marks angled 35° north of east. How fast was the Volkswagen going just before the impact?

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Hi, everyone in this practice problem. We're being asked to calculate the speed of our second student before the collision. We will have a first student riding a bicycle at a speed of five m per second. Heading north when he actually collides with another student on a bicycle heading east. The first student and his bicycle have a mass of 80 kg. And the second student and his bicycle have a mass of 60 kg. The two bicycles actually get entangled and slide through a stop leaving skid marks on the ground at an angle of 42° north of east from the point of collision. We're being asked to calculate the speed of the second student just before the collision. So the options given are a 4.39 m per second. B 6.65 m per second. C 7.41 m per second and D 8.21 m per second. So I'm going to start us off with listing all the given information in our problem statement. So M one or the mass of the first student with the bicycle is going to be 80 kg and V one or the velocity of the first student with the bicycle is going to be five m per second M2, which is the mass of the second bicycle with the cyclist is going to be kg and V2 is what we are interested at which is unknown for now the angle theta after the collision is going to be 42 degrees north of east. And what we're trying to look for is the philosophy of the second cyclist before the collision. So what we want to have to do here is to utilize the law of conservation of momentum where in the conservation of momentum, uh the momentum initial will equals to the momentum final of our defined system. What we have here is a collision of the first student and the second student. So I am going to just write down the first student is heading north. So this is going to be our first student and this is going to be our second student and they're colliding just like. So the first student going north, the second student going east, they're colliding here And then from there, they will move with an angle of 42° north of East just like. So, so what we wanna do is to actually separate the conservation of momentum. Uh One is going to be in the X direction or along the X axis and the second one is going to be in the Y direction or along the Y axis. So I am going to use the blue collar to write our first conservation of momentum in the X direction. So in the X direction, we are going to focus for the initial of the momentum the in uh the momentum of the initial before the collision, only on the second student because the first student is having ignored. So it has no sort of um components on the X direction at all. So in the X direction P I equals P F and B I is going to be M two multiplied by V two and B F is going to be M one plus M two multiplied by V F, which in this case, the V F is actually going to be half to be projected. So the V F has to be projected and that is going to be, this is going to be our velocity right here. And in the X direction, the V F then has to be multiplied by a cosine of theta just like so awesome. So next, we're going to focus on the Y direction, I'm going to change color here to red to make it easier to differentiate the two. So in the Y direction P I will still equals to P F. But in this case, we won't have any component for the initial momentum from student two. So we are only going to focus on student one. So M one multiplied by P one is actually going to equals to M one plus M two multiplied by V F multiplied by, in this case, the projection in the Y direction is going to be multiplied by sine theta just like so awesome. So now that we have two of the equations that we need, what we wanna do is to actually divide them in order for us to be easily, to easily simplify them. So I'm just gonna divide um the, I'm gonna uh abbreviate conservation of momentum by C O M. So I'm gonna defy C O M in the right direction, divide that by C O M in the X direction. So the C O M in the right direction is going to actually be M one multiplied by V one equals to M one plus M two multiplied by V F multiplied by a sign of data that will be divided. So M1 multiplied by V one is going to be divided by M2 multiplied by V two and the right one, M one plus M two multiplied by V F multiplied by assign data is going to be divided by M one plus M two multiplied by V I multiplied by cosine or V F. I mean, I'm sorry about that multiplied by V F multiplied by cosine data. So then we can simplify the right side of our equation. So the M one and M two will cancel out, the V F will also cancel out. So in this case M one P one divided by M two P two will then come out to be sin theta divided by cosine theta which will equals to the tangent of the. So now we can actually uh calculate this because then we have all the necessary information for us to find feed two. So M1 is going to be 80 kg. B one is going to be five m/s. M2 is going to be 60 kg and V two is going to be unknown. All of that is going to equals to tangent of data, which data is going to be 42 degrees just like. So, so we're going to do some rearrangement here. So V two will then equals to 80 kg multiplied by five m per second, divided by 60 kg Multiplied by tangent of 42°. And that will give us Uh two value of 7.41 meters per second. So the second student's philosophy is going to be 7.41 m per second before the collision, which will correspond to option C. So option C is going to be the answer to this particular practice problem and that'll be all for this one. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topic and they'll be it for this video. Thank you so much.

Verified Solution

Key Concepts

Conservation of Momentum

Vector Components

Collision Types