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?)
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?
3 answers
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
there was a push for .0003 seconds
then his speed was 5 m/s
so
change in momentum = 90*5 - 90*0 = 90*5