Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.

Question

please correct my answers

a.  What is the total momentum of the system before the collision? (Write your answer in terms of the mass, m.) (0.5 points)

10+8=18

b.      What is the total momentum of the system after the collision? (Write your answer in terms of the mass, m.) (0.5 points)

2m left

c.      Write puck 1's velocity after the collision in component form. (1 point)

2m

d.      What is the y-component of puck 2's velocity after the collision? (1 point)

Y=8m/2m=4m

e.      What is the x-component of puck 2's velocity after the collision? (1 point)

f.       At what angle does puck 2 move after the collision? Determine the angle and draw it on the diagram. (1 point)

g. What is the magnitude of puck 2's velocity after the collision?

(after velocity of pluck 1 is 6m/s and the angle is 25degrees)

Answer 1

First, keep in mind that momentum is a vector quantity, and is always conserved. Therefore, the momentum before and after the collision must be the same.

Before doing the details, let's establish a coordinate system: Using the typical x-y plot, let's define "right" as +x, and "left" as -x.

So--before the collision--Puck 1 has a velocity of -10, and Puck 2 has a velocity of +8. The total momentum is 8m - 10m = -2m

After the collision, the momentum is the same, but the combined mass is 2m. So the velocity (of both pucks) is -2m / 2m = -1

Based on the problem statement--with the pucks moving left and right--I would conclude there is no y component--in other words, all the action takes place along the x-axis. (No angles to calculate)