I understand momentum pretty well but what I don't understand are when the objects 'separate' ; the problem reads:
"A 90 kg man on his 2.5 kg skateboard, both traveling at 5 m/s, jumps off the board in the opposite direction at 5 m/s. What was the resultant velocity of the board?"
Could you please explain how to solve this? Thanks, I need help.
"A 90 kg man on his 2.5 kg skateboard, both traveling at 5 m/s, jumps off the board in the opposite direction at 5 m/s. What was the resultant velocity of the board?"
Could you please explain how to solve this? Thanks, I need help.
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This is the opposite of a "sticky" collision, but the principle is the same...the total momentum of the system must be the same before and after.
It might be easier to employ a moving frame of reference so that the man and his skateboard are initially at rest (0 momentum). When the man jumps off at 5 m/s backwards (-5 m/s), he acquires a momentum of:
p = mv = (90 kg)(-5 m/s) = -450 kg*m/s
Meaning the skateboard acquires a forward momentum of 450 kg*m/s (remember, this is relative to our moving reference frame). It's forward velocity is found by using p = mv
450 kg*m/s = (2.5 kg)v
v = 180 m/s
That's 180 m/s relative to the reference frame, which is itself moving forward at 5 m/s, so the skateboard's forward momentum is 185 m/s.
IMPORTANT TO NOTE: The problem doesn't specify whether the man's backwards velocity is relative to the skateboard or to the ground. It's a little confusing. If the man's velocity was relative to the GROUND rather than to the SKATEBOARD, then his backward velocity relative to the skateboard was -10 m/s, giving him a momentum of -900 kg*m/s relative to the moving reference frame. That means the skateboard's ground-relative velocity is 365 m/s...faster than the speed of sound! I went with the slightly more realistic interpretation, but either answer could be considered correct, depending on your frame of reference.
I hope that helps. Good luck!
It might be easier to employ a moving frame of reference so that the man and his skateboard are initially at rest (0 momentum). When the man jumps off at 5 m/s backwards (-5 m/s), he acquires a momentum of:
p = mv = (90 kg)(-5 m/s) = -450 kg*m/s
Meaning the skateboard acquires a forward momentum of 450 kg*m/s (remember, this is relative to our moving reference frame). It's forward velocity is found by using p = mv
450 kg*m/s = (2.5 kg)v
v = 180 m/s
That's 180 m/s relative to the reference frame, which is itself moving forward at 5 m/s, so the skateboard's forward momentum is 185 m/s.
IMPORTANT TO NOTE: The problem doesn't specify whether the man's backwards velocity is relative to the skateboard or to the ground. It's a little confusing. If the man's velocity was relative to the GROUND rather than to the SKATEBOARD, then his backward velocity relative to the skateboard was -10 m/s, giving him a momentum of -900 kg*m/s relative to the moving reference frame. That means the skateboard's ground-relative velocity is 365 m/s...faster than the speed of sound! I went with the slightly more realistic interpretation, but either answer could be considered correct, depending on your frame of reference.
I hope that helps. Good luck!