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Momentum. This questions has to be incorrect. Can't figure it out.

1. A ball weighing 520g is thrown straight up at 6.0m/s
a) What is the initial momentum of the ball in N.s




2. p=mv



=520g(1kg/1000g)(6.0m/s)
=3.12
=3.1 N.s


I don't see how this is correct because it takes kg.m.s^2 to create a Newton. There is none here.

Then we go on to

B)What is the momentum when the ball is at peak?
p=mv
=520g(1kg/1000g)(0 m/s)
=0 kg m/s

C)What is the momentum of the ball when it hits the ground?
No idea how to even go about doing C).. This one through me for a loop.

D)What is the change in momentum as the ball comes back to its original position?
P=m(Vf-Vi)
P=520g(1kg/1000g)(-6.0m/s - 6.0m/s)
=-6.24
=-6.2N.s

Any help be greatly appreciated. Thank you.
 
585
2
a) N.s = s.kg.m/s^2 = kg.m/s which are the correct units for momentum.
b) looks right
c) you already figured this out in part d
d) looks right
 
10
0
Ok this problem is interesting. For starters the units they ask for are N.s so newton*seconds. a newton is 1kg*m/s^2. So 1Kg*m/s^2 *s=1Kg*m/s which is the correct unit for momentum. So there is nothing wrong with saying a momentum is the same thing as a Newton second. (If you want me to explain that more thats fine). Moving on. you did b correctly. c. start by using kinematics to determine how far up the ball initially traveling at 6m/s with a downward acceleration of 9.81m/s^2 acting upon it, to reach its peak (final velocity=0). Then use more kinematics to determine how fast the ball is moving when it falls from zero initial velocity at a height of h (the peak height you determined) to the ground. d. When you get that velocity multiply it by the mass and that would give you the momentum on the way down. so find the difference of when the ball is at the peak and when the ball almost hits the ground. I think that's what part d is asking. Im not quite sure if it wants the change of momentum from the peak to almost hitting the ground. or from the initial toss of the ball to the ball returning to the ground.
 

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