Collisions in One Dimension On a frictionless horizontal air table, puck A (with
ID: 1474253 • Letter: C
Question
Collisions in One Dimension
On a frictionless horizontal air table, puck A (with mass 0.252 kg ) is moving toward puck B (with mass 0.368 kg), which is initially at rest. After the collision, puck A has velocity 0.116 m/s to the left, and puck B has velocity 0.654 m/s to the right.
Part A
What was the speed vAi of puck A before the collision?
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Part B
Calculate K, the change in the total kinetic energy of the system that occurs during the collision.
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Collisions in One Dimension
On a frictionless horizontal air table, puck A (with mass 0.252 kg ) is moving toward puck B (with mass 0.368 kg), which is initially at rest. After the collision, puck A has velocity 0.116 m/s to the left, and puck B has velocity 0.654 m/s to the right.
Part A
What was the speed vAi of puck A before the collision?
vAi = m/sSubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 3 attempts remaining
Part B
Calculate K, the change in the total kinetic energy of the system that occurs during the collision.
K = JSubmitHintsMy AnswersGive UpReview Part
Explanation / Answer
Solution:
Given data
puck A (with mass 0.252 kg ) is moving toward puck B (with mass 0.368 kg) which is initially at rest.
After the collision, puck A has velocity 0.116 m/s to the left,
and d puck B has velocity 0.654 m/s to the right.
a)Momentum is conserved so momentum after collision is 0.368x0.654- 0.252x0.116=0.211.
so velocity before collision is 0.211=0.252xv
v=0.211/0.252=0.845 m/s answer
Total kinetic energy before collision is 0.5x0.252x0.845^2=0.0899J.
Total kinetic energy after is 0.5x0.252x0.116^2+0.5x0.368x0.654^2=0.0786...
change in kinetic energy=0.0899-0.0786
change in kinetic energy=0.0113 L ANSWERE