On a frictionless horizontal air table, puck A (with mass 0.247 kg ) is moving t

Question

On a frictionless horizontal air table, puck A (with mass 0.247 kg ) is moving toward puck B (with mass 0.369 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 0.650 m/s to the right.

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

On a frictionless horizontal air table, puck A (with mass 0.247 kg ) is moving toward puck B (with mass 0.369 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 0.650 m/s to the right.

Explanation / Answer

mA = mass 0.247 kg

mB = mass 0.369 kg

VA = velocity 0.125 m/s

VB = velocity 0.650 m/s

mA uA + 0 = mA VA + mB VB

uA = ( mA VA + mB VB ) / mA

uA = ( 0.247 X 0.125 + 0.369 X 0.65 ) / 0.247 + 0.369

uA = 0.4394

intial KE1 = 0.5 X mA X uA2

= 0.5 X 0.247 X 0.43942

KE1 = 0.02384 J

KE2 = ( 0.5 X mA VA2 + 0.5 mB VB2 )

= ( 0.5 X 0.247 X 0.1252 + 0.5 X 0.369 X 0.652 )

KE2 = 0.07988 J

K = KE2 - KE1

K = 0.07988 - 0.02384

K = 0.05604 J

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