Consider the collision of m1=0.52g puck moving as shown in the graph, which coll
ID: 1309558 • Letter: C
Question
Consider the collision of m1=0.52g puck moving as shown in the graph, which collides with a m2=0.32g puck initially at rest. Only the track of the m1=0.52g puck, which is initially moving, is shown in the graph. Consider each point to be a picture of the puck and consider that 60 pictures ere taken each second.
a) What is the initial momentum of the m1 puck? (Write in both i and j format and as a magnitude and an angle.)
b) What is the final momentum of the m1 puck? ( Write in both i and j format and as a magnitude and an angle.)
c) What is the final momentum of the m2 puck? (Write in both i and j format and as a magnitude and an angle.)
Explanation / Answer
Part A)
SInce 60 pics are snapped each second, the straight line portion shows that there are 4 pictures taken after the initial start time. That 4/60 secs, or .0667 sec
The distance is 4 cm, so v = d/t
v = .04/.0667 = .6m/s
Momentum = mv
m = .52/1000 = .00052 kg
p = (.00052)(.6) = 3.12 X 10-4 kg m/s
Thus 3.12 X 10-4i + 0j at 0 degrees
Part B)
We have 6 time slots for pictures from 4 to 7 cm
Time = 6/60 = .1 sec
d = 3cm
v = d/t = .03/.1 = .3 m/s
p = (.00052)(.3) = 1.56 X 10-4 kgm/s
In the y, the time is still .1 sec
The distance is -1.5 cm
v = d/t = -.015/.1 = -.15 m/s
p = (.00052)(-.15) = -7.8 X 10-5 kgm/s
Thus, combining we get 1.56 X 10-4i - 7.8 X 10-5j kgm/s
The angle is from the tangent function
tan(angle) = 7.8 X 10-5/1.56 X 10-4
angle = 26.6o below the x, which is 333.4o from the axis
The net magnutide is from the pythagorean theorem
net2 = (7.8 X 10-5)2 + (1.56 X 10-4)2
net = 1.74 X 10-4kg m/s
Part C)
Since momentum must be conserved, the m2 momentum must be
1.56 X 10-4i + 7.8 X 10-5j kgm/s
The angle has to be 26.6ofrom the axis
The magnitude will be the same as part B
That is net = 1.74 X 10-4 kg m/s