Question

On a frictionless horizontal air table, puck A (with mass 0.255kg ) is moving toward puck B (with mass 0.368kg ), which is initially at rest. After the collision, puck A has velocity 0.125m/s to the left, and puck B has velocity 0.646m/s to the right.

A. What was the speed vAi of puck A before the collision?

vAi = m/s B.Calculate ?K, the change in the total kinetic energy of the system that occurs during the collision. ?K = J

Answer 1

momentum will be conserved.

so 0.255*v=0.255*(-0.125)+0.368*0.646

v=0.807 m/s

b)kinetic energy lost=initial kinetic energy-fianl kinetic enrgy

=0.5*(0.255*v^2-0.255*0.125^2-0.368*0.646^2)=0.0043108 J

Answer 2

The same question was resolved before with different figures. Please input your figure to get your answer. Let me know if you face any issue. Please rate if I succeeded in helping you.

On a frictionless horizontal air table, puck A (with mass 0.255 kg) is moving toward puck B (with mass 0.373 kg), which is initially at rest. After the collision, puck A has velocity 0.115 m/s to the left, and puck B has velocity 0.655 m/s to the right

old question

A. What was the speed v_Ai of puck A before the collision

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

Answer

Before the collision:
mv + MV = (.255kg)(v) + (.373kg)(0m/s) = v(.255kg)

KE1 = 1/2(.255kg)(v)^2 + 1/2(.373kg)(0m/s)^2 = 1/2(.255kg)(v)^2

After the collision:
mv + MV = (.255kg)(-.124m/s) + (.373kg)(.649m/s)

KE2 = 1/2(.255kg)(-.124m/s)^2 + 1/2(.373kg)(.649m/s)^2

But momentum is conserved, so:
(.255kg)(-.124m/s) + (.373kg)(.649m/s) = v(.255kg)