🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 28 Easy Difficulty # Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancingblow. Asteroid$A$, which was initially traveling at$40.0 \mathrm{m} / \mathrm{s},$is deflected$30.0^{\circ}$from its original direction, while asteroid$B$travels at$45.0^{\circ}$to the original direction of$A(\text { Fig. } .8 .36) .$(a) Find the speed of each asteroid after the collision. (b) What fraction of the original kinetic energy of asteroid A dissipates during this collision? ### Answer ## (a)$v_{A}=29.3 \mathrm{m} / \mathrm{s} \quad-\quad v_{B}=20.7\mathrm{m} / \mathrm{s}$(b)$f=19.6 \%\$

#### Topics

Moment, Impulse, and Collisions

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

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### Video Transcript

in this question, Asteroid A was moving to the right with a velocity of 40 m per second. When it collides with asteroid, be after that collision. Asteroid A is moving with a new velocity V A with an angle off 30 degrees with respect tweets Traditional trajectory. Andi Asteroid B is now moving with a velocity V. Be making an angle off 45 degrees with the original trajectory off the asteroid A. In the first item, we have to find the speed off each asteroid after the collision. That is, we have to find V A M v B. So for that you have to remember that the net momentum is conserved both before and after the collision. So the net momentum pew net before is equals the net momentum after pew a net. Now we have to deal with two dimensions because we have two directions off movement here. The movement off Asteroid A happens to the right and happens up on the movement off Asteroid be happens to the right and downwards the reform. We are working with two dimensions. So in this question, my reference frame will be this one. Everything that is pointing upwards will be positive and everything that is pointing to the right will also be positive. Let me call this horizontal axis the X axis and the vertical one, the Y axis. Okay, so what happens is that the law of conservation of momentum must hold in both off these axes. So these equation that I wrote right here actually is equivalent to two equations, one for each direction. So the net momentum in the X direction before the collision must be equals to the net momentum in the X direction after the collision and at the same time we must have. That's the net momentum in the Y direction before the collision should be equals to the net momentum in the Y direction after the collision. So we actually have two equations toe work. And now let us use these equations. So we know that to evaluate the momentum, we have to multiply the mass by the velocity. Then to evaluate the momentum in the X direction, we multiply the mass by the velocity in the X direction. So the first equality goes as follows. So in the X direction before the collision we had on Lee the asteroid a which has a mass m A which is equal to the mass off asteroid be. But we had on Lee the asteroid a moving to the right with the velocity off 40 m per second. Then its momentum waas the mass off asteroid a times the initial velocity off 40 when it was moving to the right. So we have a positive sign and then after the collision, what we have is both asteroids moving with some velocity in the X direction that we don't know. So what happens is that after the collision, the net momentum is given by the mass off the asteroid A times the velocity off Asteroid eight after the collision in the X direction, plus the mass off Asteroid B which is equal to the mass off asteroid A. So I'm writing a again times the velocity off asteroid be in the X direction after the collision. Then we can simplify the masses. That is, we can divide by m a, both the left hand side on the right hand side and we end up with the following v A. In the X direction plus VB in the X direction is the cause to 40 now, what is the relation off v A in the X direction with V A and VB in the X direction with Phoebe. For that, we have to decompose vectors as follows. So here is our reference frame. This is our X axis and this is our Y axis. Now let me draw the velocity V A after the collision. So it is something like that. This is V A. We know that the angle that v a makes with the horizontal so it makes with our X axis is 30 degrees than here. We haven't angle off 30 degrees. Now we want to evaluate what is the X component off the velocity v a. What to do? Well, it's easy if we draw it. So this is the X component off v A. On this is the white component which will also use. So here we have V A. Why component. And here we have V a X component. Now note the following In order to deter mined the relation between V a X component and V A, we can use the co sign off 30 because of the following. In this context, the co sign off 30 is given by the address and side off triangle, which is V a X component divided by the high party news which is V a. The Reform V a X component is equal to the hypotenuse v a times that co sign off 30 degrees. And now we can use this results in this expression to get the following V eight times the co sign off 30 degrees plus Now VB also follows the same idea. So we have vb times. The co sign off 45 degrees is equals to 40 and this is how you know up to this point now we go to the Y direction in the Y direction we have the following Before the collision, there was no movement at all in the vertical axis, so the net momentum before the collision was equals to zero. But after the collision, both asteroids are also moving in the Y direction. So we have m a times V a y direction plus M b which is equals to m A. So I'm writing m a times vb in the Y direction again. We can divide both the left hand side and the right hand side by m a to get the following v a in the Y direction plus VB in the Y direction is equal to zero Now what is the relation between V A in the Y direction and V A and VB in the right direction and vb For that we can do the following again. Take a look at this triangle Now we want a relation between V A Y component and V A. As you can see, we can use the sign off 30 degrees because the sign off 30 degrees in this context is given by the following. It is the opposite side off the triangle, which is V a Y component divided by the hypotenuse which is v A to reform its true that V A y component is equals to v a times this sign off 30 degrees. So here we have V a times the sign off 30 degrees minus vb times this sign off 45 degrees and this is equals to zero. Now you might want to know why there is a negative sign here. And the negative sign here is because, as you can see, the Y component off VB will point downwards in the negative direction off our vertical axis. So this is why I have this minus sign here. Then using this equation, you can bet remind the following V a times The sign off 30 degrees is equals to vb times. They sign off 45 degrees. The Reform V A is equals to vb times the sign off 45 degrees divided by the sine off 30 degrees. And now we can use this result these equations in order to finally determine what is VB. So by using these results, we get the following. Now we have vb times the sign off 45 degrees, divided by the sine off 30 degrees times the co sign off 30 degrees plus vb times. The co sign off 45 degrees is equals to 40. Then we can factor V B to get following vb times this sign off 45 degrees times the co sign off 30 degrees divided by the sine off 30 degrees, plus the co sign off 45 degrees is a close to 40 and finally, VB is equals to the sign off 45 degrees times the co sign off 30 degrees, divided by the sine off 30 degrees, plus the co sign off 45 degrees. Now those are dividing 40. This results in a speed VP off approximately 20.7 m per second. Now that we know VB, we can use this expression to Dr Mind V A. By doing that, to get the following V A is equals to 20 0.7 times the sign off 45 degrees divided by the sine off 30 degrees. By doing that, we get va being approximately 29.3 m per second. And this is the answer to the first item off this question. Now, for the second item, we will need the values off the speeds, so I will keep them here in the second item, we want to know what fraction off the original kinetic energy off Asteroid A is dissipated in the collision. So for that, we can compare the kinetic energy off the system before the collision which was composed off the kinetic energy off asteroid A only with the kinetic energy off the system after the collision. So before the collision, let me call this E B. So the kinetic energy before the collision is given by the kinetic energy off Asteroid A only so one half off the massive asteroid a times the square off its speed. So 40 squared after the collision. The kinetic energy is given by the some off the kinetic energy off Asteroid A on asteroid. Be so after the collision, we have one half off a times the speed off asteroid a 29.3 squared after the collision, plus one half off the mass off Asteroid B, which is equal to the mass off asteroid eight times the speed off Asteroid B, which is 20.7 squared after the collision. Okay, now we can calculate what is the variation in the kinetic energy. By doing that, we got the following. The variation in the kinetic energy is given by the kinetic energy after the collision. So one half m a factoring this times 29.3 squared plus 20.7 squared minus one half m a times 40 squared. This is a variation in the kinetic energy. Then we can divide the variation in the kinetic energy by the initial kinetic energy E K. And this is a fraction that we want to evaluate. This fraction is given by one half off M A times 29.3 squared plus 20.7 squared minus 40 squared, divided by one half off M a times 40 squared. We can simplify the factors off one half and m a. And then by performing the calculation, we get the fraction, which is approximately minus 0.196 which is equivalent to minus 19.6%. So the conclusion is that 19.6% off the initial kinetic energy was dissipated during the collision. We know that it was dissipated because we have this minus sign here on this is the answer to this question.

Brazilian Center for Research in Physics

#### Topics

Moment, Impulse, and Collisions

##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

Lectures

Join Bootcamp