On a frictionless, horizontal air table, puck A (with mass 0.250kg) is moving toward puck B (with mass 0.350kg), which is initially at rest. After the collision, puck A has a velocity of 0.120m/sto the left, and puck B has a velocity of 0.650m/s to the right. (a) What was the speed of puck A before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Short Answer

Expert verified

(a) The speed of puck A before collision is 0.79m/s.

(b) The change in kinetic energy is0.002J

Step by step solution

01

Step 1:Given in the question:

Mass of puck Ais mA=0.250kg

Mass of puck BismB=0.350kg

Velocity of puck A is vA=0.120m/s

Velocity of puck B is vB=0.650m/s.

02

 Step 2: Law of conservation of momentum:

When two bodies collide each other then the total momentum before collision is equal to the total momentum after collision.

03

Step 3:(a) Speed of puck A before collision:

The speed of puck A before collision is calculated as follows:

PAi+PBi=PAf+PBfmAvAi+0=mAvAf+mBvBf

vAi=vAf+mBmAvBf=-0.120+0.3500.250×0.650=0.79m/s

Hence,the speed of puck A before collision is0.79m/s

04

(b) The change in the kinetic energy of the system:

The change in the kinetic energy is calculated as follows:

ΔK=Kf-Ki=12mAvAf2+12mBvBf2-12mAvAi2=12×0.25×0.122+0.35×0.652-0.25×0.792=0.002J

Hence, the change in kinetic energy is 0.002J.

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