Question

I got Part A but need help with Part B thanks!

Head-on Collision

A blue puck with a mass of 3.60×10⁻²  kg , sliding with a velocity of 0.220  m/s North on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass m, initially at rest. After the collision, the velocity of the blue puck is 4.0×10⁻²  m/s North.

Part A

Find the momentum of the red puck after the collision. Give the horizontal component with North positive.

Give the horizontal component of momentum with North positive.

6.48×10−3

kg⋅m/s

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Correct

Part B

Find the kinetic energy of the red puck after the collision.

3.24•10−7	J

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Answer 1

The total energy of blue puck before the collision is:

KEi(b) = 1/2 m vi^2 = 1/2 x 3.6 x 10^-2 x 0.22 x 0.22 = 8.7 x 10^-4 J

The KE of blue after the collision

KEf(b) = 1/2 m vf^2 = 1/2 x 3.6 x 10^-2 x 0.04 x 0.04 = 2.9 x 10^-5 J

$\Delta$ KE = 8.7 x 10^-4 - 2.9 x 10^-5 = 8.41 x 10^-4 J

$\Delta$ KE = 8.41 x 10^-4 J

from energy conservation the energy calculated above should be equal to the KE of red puck.

KE(red) = 8.41 x 10^-4 J