Question

On a frictionless, horizontal air table, puck A (with mass 0.250 kg ) is moving toward puck B (with mass 0.400 kg ), which is initially at rest. After the collision, puck A has a velocity of 0.150 m/s to the left, and puck B has a velocity of 0.620 m/s to the right.

a.What was the speed of puck A before the collision?

b. Calculate the change in the total kinetic energy of the system that occurs during the collision.

Answer 1

   mass of A 0.25kg , m1, u1 = ?
                           m2 = 0.4 kg, u2 = 0 m/s
        after collision
           v1 = 0.150 m/s, v2 = 0.620 m/s

  

   coservation of momentum

a)
m1u1 +m2u2 = m1v1 +m2v2
u1 = (- m1v1 +m2v2 )/ m1 = ( - 0.25 *0.15 +0.4*0.62) / 0.25
               = 0.842 m/s

b) change in kinetic energy
  

       1/2 m1u1^2 + 1/2 m2u2^2 = 1/2 m1 v1^2 + 1/2 m2v2^2
      
       1/2 m1u1^2 + 0 = -1/2 m1 v1^2 + 1/2 m2v2^2
                1/2 *0.25*0.842^2 = -1/2*0.25*0.15^2 + 1/2 0.4* 0.62^2
                                  = 0.0145525 J