A man weighing 730 N and a woman weighing 400 N have the same momentum. What is

Question

A man weighing 730 N and a woman weighing 400 N have the same momentum. What is the ratio of the man's kinetic energy K_m to that of the woman K_w?

On a frictionless horizontal air table, puck A (with mass 0.252 kg) is moving toward puck B (with mass 0.367 kg), which is initially at rest. After the collision, puck A has velocity 0.121 m/s to the left, and puck B has velocity 0.651 m/s to the right.

b) Calculate DeltaK, the change in the total kinetic energy of the system that occurs during the collision.

Explanation / Answer

Momentum, p is
p = mv

Kinetic energy, KE is
KE = .5*mv^2

The Kinetic energy can be written
KE= .5*p*v

KE1/KE2 = v1/v2 (since they both have the same momentum)

find v1/v2 by dividing the momentums
p1/p2 = m1v1/(m2v2) = 1

you should also know that weight is equal to mass times gravity or

W = mg

substituting this in gives

1 = W1*v1/(W2*v2)

so
v1/v2 = W2/W1

so
KE1/KE2 = W2/W1 = 400/730 = 0.54

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