Remember that when dealing with collisions in 2-dimensions, momentum is conserve
ID: 1324935 • Letter: R
Question
Remember that when dealing with collisions in 2-dimensions, momentum is conserved in each dimension.
Two pucks are sliding on a frictionless tabletop. Block A (mass 4.95 kg) is moving to the East (+x direction) at 2.70 m/s. Block B (mass 5.94 kg) is moving to the North (+y direction) at 3.24 m/s. Assume the system to be both Block A and Block B. (Note: It could be useful to make a drawing of the situation.)
What is the x-component of the total momentum of the system before the collision?
What is the y-component of the total momentum of the system before the collision?
What is the x-component of the total momentum of the system after the collision?
What is the y-component of the total momentum of the system after the collision?
Explanation / Answer
since both the balls are moving at right angles to each other
momentum along east Px = mv
Px = 4.95 * 2.7 = 13.365 kgm/s----------<<<<<<Answer to PArt A
momentum along north = Py = mu
Py = 5.94 * 3.24 = 19.24 kgm/s------<<<<<<<Answer to part B
so
net momentum p^2 = px^2 + py^2
P^2 = 13.365 ^2 +19.242^2
P^2 = 549
P = MV = 23.43
speed V = 23.43/(4.95 + 5.94)
V = 2.1515 m/s ----------------answer to part C and D