🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Oh no! Our educators are currently working hard solving this question. In the meantime, our AI Tutor recommends this similar expert step-by-step video covering the same topics. Numerade Educator ### Problem 89 Easy Difficulty # Two asteroids with masses$m_{A}$and$m_{B}$are moving with velocities$\vec{v}_{A}$and$\vec{v}_{B}$with respect to an astronomer in a space vehicle. (a) Show that the total kinetic energy as measured by the astronomer is$$K=\frac{1}{2} M v_{\mathrm{cm}}^{2}+\frac{1}{2}\left(m_{A} v_{A}^{\prime 2}+m_{B} v_{B}^{\prime 2}\right)$$with$\vec{v}_{\mathrm{cm}}$and$M$defined as in Section$8.5, \vec{v}_{\mathrm{A}}^{\prime}=\vec{v}_{\mathrm{A}}-\overrightarrow{\boldsymbol{v}}_{\mathrm{cm}},$and$\vec{v}_{B}^{\prime}=\vec{b}_{B}-\vec{b}_{\mathrm{cm}},$In this expression the total kinetic energy of the two asteroids is the energy associated with their center of mass plus the energy associated with the internal motion relative to the center of mass. (b) If the asteroids collide, what is the minimum possible kinetic energy they can have after the collision, as measured by the astronomer? Explain. ### Answer ## a.$K=\frac{1}{2}\left(m_{A}+m_{B}\right) v_{\mathrm{em}}^{2}+\frac{1}{2}\left(m_{A} v_{A}^{\prime 2}+m_{B} v_{B}^{\prime 2}\right)+\left(m_{A} \overrightarrow{\boldsymbol{v}}_{A}^{\prime}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}^{\prime}\right) \cdot \overrightarrow{\boldsymbol{v}}_{\mathrm{cm}}$b.$\frac{1}{2} M v_{\mathrm{cm}}^{2}\$

#### Topics

Moment, Impulse, and Collisions

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##### Andy C.

University of Michigan - Ann Arbor

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

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### Video Transcript

{'transcript': "problem. 8.31. We have two asteroids colliding. They're the same mass, which is very convenient. The initial speed of this is Ah, 40 meters per second and then they go off and two directions whose angles from the initial direction of motion or given and let's call these angles stated a and they to be. And while we're at it, called this the positive extraction and this the positive direction. So because we're not told of any forces acting external, too, the asteroids or any forces that there are are a negligible because this is the asteroid belt where everything is very far apart. For the most part, the momentum in both of these directions is going to be conserved. So that's right down what we have for the extraction first. So em, which has no subscript because they both have the same mass imes the A initial holes. And now we're doing the X component. So we're going to want the co signs of these angles sign, uh, a plus coasts. I enough, they have B. We'll note that all the ends cancel out here now in the white direction. There's initially zero momentum because this asteroid is only moving and over calling the extraction. So that zero it's gonna be equal to m sign. Say that a Oh, my apologies. Um, yeah, We, uh we do want the speeds very important not to forget that. What sign? It's even more important. If you do forget that. Catch that before it's too late. So am I be a final sign of state, eh? Now, this is going to be minus because B is moving in the negative. But the white component of these velocity is in the negative. Wider action. That's, um, e sign, you know, be so all of the EMS cancel out and we're left with a system of two linear equations and two unknowns, then being the final speeds of A and B E final. And so there, you know you can take whatever way you like for solving this type of thing where you have a system of equations. It's a few ways of doing it and then plugging in the numbers you find a speed of a after. The collusion is 29.3 meters per second and the speed of be after the collision. Just 20.7 is 27. So now that we know how fast they're going after the clue, we can start thinking about would fraction or percentage of the kinetic energy is lost? So you know, the percent changing something is the change divided by its initial value. So this will be, if I'm old minus a initial divided by Okay, initial. And then this works out to be the ratio of the final and initial kinetic energies minus one. Excuse me, we a claim ALS. Where be final and all of the 1/2 ends they're going to cancel out, which is a little bit easier, mostly just writing down so that it it's clear where we get this from Take 1/2 of them's out. And so it's just the sum of the final velocity squared, divided by. And then you take one off of that and now this ends up being zero point 80 for So the way you're thinking about this is this is how much energy is left from the initial stayed at in subtracting one from that tells us what the what. The percent changes and this is a course. Oh, it's negative, or about about 20%. His loss"}

Cornell University

#### Topics

Moment, Impulse, and Collisions

##### Andy C.

University of Michigan - Ann Arbor

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp