Find the final velocities of V1 and V2 given these two equations, i know the ans

Question

Find the final velocities of V1 and V2 given these two equations, i know the answer to the question so please show the work. equation 1 [ 2mV1 + mV2 = 2mV ], equation 2 [ (1/2)(2m)V1^2 + (1/2)mV2^2 = (1/2)(2m)V^2 ] Please show the work to get to the answer. Answer V1= V/3 and V2= 4V/3

Explanation / Answer

Having said that, they're definitely doable, if you understand the concepts. The keys here are momentum and energy (both kinetic and potential). Energy is conserved, but in this situation where they stick together, momentum really isn't, so let's just talk energy. First thing, as always with physics problems, is to draw a large, clear, labeled diagram so you know what is where and what's happening. I won't bother do to that and then scan and post the image, but I'm drawing on here and you should draw one there! Immediately before the collision, sphere #2 has zero kinetic energy, because it is stationary. we can also consider that it has zero potential energy, because we can define our zero point wherever we like, and it might as well be at the 'rest height' of the spheres. Sphere #1 has mass m (both spheres do). Its kinetic energy just before the collision will be the same as the potential energy it had when raised to point A. That will be given by PE = mgh, where h is the height of A above the rest position. You're asked to calculate the height at A in terms of the velocity just before the collision, plus g. You can set KE(bottom)=PE(top): 1/2mv^2 = mgh The m on both sides cancels and it's just a matter of rearranging that equation to make h the subject. You should get h_A = v^2/(2g) (put the little subscripts on the v that the teacher used) Right, that's part a) answered. For part b), you know that the KE just before the collision was 1/2mv^2. Just after the collision, energy is conserved, but the mass has doubled. I'll use v1 for before and v2 for after the collision: 1/2mv1^2 = 1/2(2m)v2^2 ---> 1/2mv1^2 = mv2^2 (m cancels on both sides, so while the question asks for it in terms of m, it's not needed) v1^2/2 = v2^2 (flip sides and take the square root of both sides: v2 = v1/(sqrt2) For part c), you just do the reverse of part a): comvert KE(bottom) to PE(top) 1/2(2m)v2^2 = (2m)gh (cancel 2m on both sides and rearrange) h_C = v2^2/(2g) You're asked for it in terms of h_A, though, which you calculated in part a). You need to substitute in the answer from part b) to express v2 in terms of v1, then use the expression for h_A you calculated: h_C = v1^2/4g but h_A = v1^2/2g, so h_C = h_A/2

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