Question

On a frictionless, horizontal air table, puck A (with mass 0.250 kg ) is moving toward puck B (with mass 0.350 kg ), that is initially at rest. After the collision, puck A has a velocity of 0.121 m/s to the left, and puck B has velocity 0.650 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

here,
mass of puck a, ma = 0.250 kg
mass of puck b, mb = 0.350 kg

after collision velocity of puck a, va' = -0.121 m/s
after collision velocity of puck b, vb' = 0.650 m/s

from conservation of momentum, before collision = after collision

ma*va + 0 = ma*va' + mb*vb'

solving for velocity of puck a before collision, va

va = (ma*va' + mb*vb')/ma
va = (0.250*(-0.121) + 0.650*0.350)/0.250
va = 0.789 m/s

Part B:
Kinetic energy before collision, Kei = 0.5*ma*va^2 = 0.5*0.250*0.789^2 = 0.078 J

Kinetic energy after collision, Kef = 0.5*ma*va'^2 + 0.5*mb*vb' = 0.5*0.250*(-0.121)^2 + 0.5*0.350*0.650^2 = 0.076 J

Net Change in KE, KE = 0.078 * 0.076 = 0.002 J