# A.) What was the speed vAi of puck A before the collision? B.) Calculate ΔK, the change in the total kinetic energy of the system?

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On a frictionless horizontal air table, puck A (with mass 0.253 kg ) is moving toward puck B (with mass 0.366 kg ), which is initially at rest. After the collision, puck A has velocity 0.118 m/s to the left, and puck B has velocity 0.649 m/s to the right.

• For A, initial x momentum = 0.03 * 7.79 = 0.2337

Final x momentum = 0.03 * vA * cos 65

Final y momentum = 0.03 * vA * sin 65

For B, final x momentum = 0.06 * vB * cos 37

Final y momentum = -0.06 * vB * sin 37

0.03 * vA * cos 65 + 0.06 * vB * cos 37 = 0.2337

0.03 * vA * sin 65 + -0.06 * vB * sin 37 = 0

0.03 * vA * sin 65 = 0.06 * vB * sin 37

vA = 2 * vB * sin 65 ÷ sin 37

0.03 * (2 * vB * sin 65 ÷ sin 37) * cos 65 + 0.06 * vB * cos 37 = 0.2337

vB * (0.6 * sin 65 ÷ sin 37 cos 65 + 0.06 * cos 37) = 0.2337

vB = 0.2337÷ [(0.6 * sin 65 ÷ sin 37 * cos 65 + 0.06 * cos 37)]

vB = 0.2337 ÷ 0.42948519

This is approximately 0.544 m/s.

vA = 2 * (0.2337 ÷ 0.42948519) * sin 65 ÷ sin 37

This is approximately 1.63 m/s.

Check:

0.03 * 1.63* cos 65 + 0.06 * 0.544 * cos 37 ≈ 0.2327

Since I rounded the velocities, I believe that this proves that they are correct.

• Let v = velocity of A before collision to the right

Momentum before collision = 0.253v

Momentum after collision = (0.649)(0.366) – (0.253)(0.118)

= 0.20768

Therefore 0.253v = 0.20768

v = 0.20768/0.253

= 0.8208695652

= 0. 821 m/s

KE of A before collision = 1/2(0.253)(0.821)^2

= 0.0853 J

KE of A after collision = 1/2(0.253)(0.118)^2

= 0.00176 J

KE of B after collision = 1/2(0.366)(0.649)^2

= 0.0771 J

Total KE after collision = 0.0771 + 0.00176

= 0.07886

ΔK = 0.0853 – 0.07886

= 6.44 x 10^-3 J ….[ = 0.00644 J]

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