A ball of mass M collides head-on and in elastically with a ball of mass 3M. Bef

Question

A ball of mass M collides head-on and in elastically with a ball of mass 3M. Before the collision, the hall of mass M is moving to the right at speed v_0 and the ball of mass 3M is moving to the left, at the same speed v_0. After the collision, the ball of mass 3M is moving to the left at speed v_0/2. What fraction of the initial kinetic energy was lost in the collision? What fraction of the initial kinetic energy is lost in the collision? 1. 1 2. 0 3. 1/6 4. 1/5 5. 2/3 6. 1/4 7. 1/3 8. 3/4 9. 1/2 10. 5/6

Explanation / Answer

Using law of conservation of momentum

Momentum before collision = momentum after collision

M*vo -3Mvo = -3M*(Vo/2) + (M*v)

Vo-3Vo = (-3*Vo/2) + v

-2*Vo = (-3*Vo/2)+v

v = (3*Vo/2) - 2Vo

v = Vo(1.5-2) = -0.5*vo = -vo/2

initial kinetic energy is Ki = 0.5*M*Vo^2 + (0.5*3M*vo^2) = 2*M*vo^2

Final kinetic energy is Kf = (0.5*M*(vo/2)^2) + (0.5*3M*(vo/2)^2)

Kf = 2*M*(Vo/2)^2 = M*Vo^2/2

change in kinetic energy is Kf-Ki = (0.5*M*vo^2)-(2*M*vo^2) = -(3/2)*M*vo^2

required fraction is (Kf-Ki)/ki = (-3/2)*M*vo^2 / (2*M*vo^2) = 3/4

So the answer is 8) 3/4

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