- #1
Underdog_85
- 10
- 0
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.
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.