Two students (90.0 kg and 60.0 kg) on roller skates face-to-face push against each other. The 90.0 kg student moves at 5.0 m/s just after their hands lose contact.
I figured out part A (final velocity of 60.0 kg student) to be 7.5 m/s backward. I need help with part B.
(b) What average force was exerted on each student if they were in contact for 0.0003 s?
force = change in momentum / change in time
90*5/.0003 in Newtons
by the way .0003 is a very short pushing time, impractical I think. (typo?)
You're right, it is small, but it's not a typo. How is the change in momentum 90*5?
The 90 kg student was standing still, momentum of zero.
there was a push for .0003 seconds
then his speed was 5 m/s
so
change in momentum = 90*5 - 90*0 = 90*5
To determine the average force exerted on each student, we can use the principle of conservation of momentum.
The formula for momentum is:
Momentum = mass * velocity
According to the question, the 90.0 kg student has a final velocity of 5.0 m/s after their hands lose contact, and the 60.0 kg student has a final velocity of 7.5 m/s backward.
Using the principle of conservation of momentum, we know that the initial momentum of the two students should be equal to the final momentum.
Initial momentum of the system = Final momentum of the system
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 * final_velocity1) + (mass2 * final_velocity2)
Let's plug in the values:
(90.0 kg * 0 m/s) + (60.0 kg * 0 m/s) = (90.0 kg * 5.0 m/s) + (60.0 kg * 7.5 m/s)
0 + 0 = 450.0 kg * m/s + 450.0 kg * m/s
0 = 900.0 kg * m/s
Since the initial momentum is zero, this implies that the final momentum is also zero.
Now, we can calculate the average force exerted on each student using the formula:
Force = (Change in momentum) / (Time)
Since the change in momentum is zero, the average force exerted on each student is also zero.
Therefore, the answer to part B is that the average force exerted on each student is 0 Newtons.