I\'ll write the problem out as is, but my question is at the end. The mass of th

Question

I'll write the problem out as is, but my question is at the end.
The mass of the blue puck is 20% greater than the mass of the green one. Before colliding, the pucks approach each other with equal and opposite momentum, and the green puck has an initial speed of 10 m/s. Find the speeds of the pucks after the collision if half the kinetic energy is lost during the collision.

so we have total momentum = 0
m1v1 = m2v2
m1 (10m/s) = 1.2m1 v2
v2= 8.33 m/s the initial speed of the blue puck

KE before collision
1/2 m1 (10)^2 + 1/2 (1.2m1)(8.33 m/s)^2
1/2m1 (100) + 1/2 (1.2m1)(69.3889)
1/2m1 (100) + 1/2m1(83.26668)
1/2m (183 m^2/S^2

KE final
1/2 m v^2 + 1/2 (1.2m) v^2 = 1/2 [1/2m (183.3 m^2/s^20]
v^2 + 1.2 v2 = 91.7 m^2/s^2 eq 1
Mgreen vfinal = 1.2mvblue final or vgf=1.2vbf eq2

up to here I understand what is going on. I'm supposed to solve 1 and 2 simultaneously (HUH?) to obtain vgf = 7.1 m/s and vbf = 5.9 m/s
How do I do that? Thanks!

Explanation / Answer

Let m1 = mass of green puck m2 = mass of blue puck u1 = initial speed of green puck u2 = initial speed of blue puck m2 = m1 + 20% of m1 m2 = 1.2 m1--------------(1) u1 = 10.0 m/s------------(2) Before collision, the pucks momenta are equal and opposite. Therefore total momentum = 0 m1 u1 = m2 u2 m1 * 10.0 = 1.2 m1 * u2------------[using (1) and (2)] 10.0 = 1.2 * u2 u2 = 10.0/1.2 u2 = 8.33 m/s---------(3) Total KE before collision = (1/2)m1 u1^2 + (1/2)m2 u2^2 = (1/2)m1 * 10^2 + (1/2) * 1.2 m1 * 8.33^2 = 50 m1 + 41.67 m1 = 91.66 m1 Half KE is converted into internal energy. Therefore total KE after collision = (91.66/2)m1 = 45.83 m1-----(4) Let speed of green puck after collision = v1 speed of blue puck after collision = v2 By conservation of momentum, m1 v1 = m2 v2 m1 v1 = 1.2 m1 v2 v1 = 1.2 v2-----------------(5) Total KE after collision = (1/2)m1 v1^2 + (1/2)m2 v2^2 Using (1), (4), (5) in the above, 45.83 m1 = (1/2)m1 (1.2 v2)^2 + (1/2) * 1.2 m1 * v2^2 45.83= (1/2)(1.2 v2)^2 + (1/2) * 1.2 * v2^2 45.83= 0.72 v2^2 + 0.6 v2^2 45.83= 1.32 v2^2 v2^2 = 45.83/1.32 = 34.72 v2 = sqrt(34.72) = 5.89 m/s From (5), v1 = 1.2 * v2 = 1.2 * 5.89 = 7.07 m/s Ans: Speed of green puck = 7.07 m/s Speed of blue puck = 5.89 m/s

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