Question

16. On a frictionless horizontal air table, puck A (with mass 0.255 kg ) is moving toward puck B (with mass 0.375 kg ), which is initially at rest. After the collision, puck A has velocity 0.125 m/s to the left, and puck B has velocity 0.649 m/s to the right.

Part A: What was the speed vAi of puck A before the collision?

Part B: Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.

Answer 1

Let us consider the right side direction as positive and the left side direction as negative.

Mass of puck A = m₁ = 0.255 kg

Mass of puck B = m₂ = 0.375 kg

Initial velocity of puck A = V₁

Initial velocity of puck B = V₂ = 0 m/s (At rest)

Final velocity of puck A = V₃ = -0.125 m/s (Negative as it directed to the left)

Final velocity of puck B = V₄ = 0.649 m/s

By conservation of linear momentum,

m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄

(0.255)V₁ + (0.375)(0) = (0.255)(-0.125) + (0.375)(0.649)

V₁ = 0.829 m/s

Total kinetic energy of the system before the collision = KE₁

KE₁ = m₁V₁²/2 + m₂V₂²/2

KE₁ = (0.255)(0.829)²/2 + (0.375)(0)²/2

KE₁ = 8.76 x 10^-2 J

Total kinetic energy of the system after the collision = KE₂

KE₂ = m₁V₃²/2 + m₂V₄²/2

KE₂ = (0.255)(-0.125)²/2 + (0.375)(0.649)²/2

KE₂ = 8.09 x 10^-2 J

Change in total kinetic energy of the system = KE

KE = KE₂ - KE₁

KE = 8.09x10^-2 - 8.76x10^-2

KE = -6.7 x 10^-3 J

Negative as kinetic energy is lost in the collision.

A) Speed of puck A before the collision = 0.829 m/s

B) Change in total kinetic energy of the system during the collision = -6.7 x 10^-3 J