On a frictionless air track, a 0.5000 kg-glider moving to the right at 2.950 m/s

Question

On a frictionless air track, a 0.5000 kg-glider moving to the right at 2.950 m/s collides with a 0.4000 kg glider moving to the left at 1.950 m/s. The collision is elastic. Assume the right direction is the positive direction of motion . (a) What is the velocity of the 0.5000 kg-glider after the collision? (b) What is the velocity of the 0.4000 kg-glider after the collision? (c)Using the velocities of parts (a) and (b), compute the total kinetic energy of the system after the collision. Do not round off during intermediate computations. (d) Using the initial velocities before the collision, compute the total kinetic energy of the system before the collision. Do not round off during intermediate computations. (e) (Are the answer to parts (c) and (d) equal? Explain.

Explanation / Answer

A) let m1 = 0.5 kg and u1 = 2.95 m/s

m2 = 0.4 kg and u2 = -1.95 m/s

then velocity of 0.5 kg after the collision is v1 = [(m1-m2)*u1/(m1+m2)] - [(2*m2*u2)/(m1+m2)]

v1 = ((0.5-0.4)*2.95/(0.5+0.4))- ((2*0.4*1.95)/(0.5+0.4)) = -1.405 m/s

1.405 m/s left side

B) v2 = [2*m1*u1/(m1+m2)] - [(m2-m1)*u2/(m1+m2)]

v2 = (2*0.5*2.95/(0.5+0.4)) - ((0.4-0.5)*1.95/(0.4+0.5)) = 3.494 m/s

right side

C) KEf = 0.5*m1*v1^2 + 0.5*m2*v2^2

KEf = (0.5*0.5*1.405*1.405) +(0.5*0.4*3.494*3.494) = 2.94 J

D) KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEi = (0.5*0.5*2.95*2.95) +(0.5*0.4*1.95*1.95) = 2.94 J

E) yes answers to parts (c) and (d) are equal

since in elastic collision KE after collision = KE before collision

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