Question

Joe applies a force of 20 N for 1/20 sec. to a soccer ball. What is its change in momentum? If the soccer ball's mass is 0.5 kg, how fast is it going after he kicks it?

Thank you and I appreciate the help!

Answer 1

@shamim @satellite73 @SolomonZelman @dan815 @TheSmartOne @zepdrix @Loser66 @Luigi0210 @robtobey @AlexandervonHumboldt2 @uri @Mehek14 @sleepyjess @ikram002p @Miracrown

Answer 2

@Hero@dan815 @TheSmartOne @Pooja195@Luigi0210 @AlexandervonHumboldt2 @triciaal@johnweldon1993 @satellite73 @zepdrix

Answer 3

@amistre64 @Astrophysics @dan815 @satellite73 @zepdrix

Answer 4

@TheSmartOne @Luigi0210 @AlexandervonHumboldt2 @Compassionate @mathmale @mathmate @jhonyy9 @sleepyjess @Miracrown @ShadowLegendX @UsukiDoll @just_one_last_goodbye @Astrophysics

Answer 5

@dan815 @TheSmartOne @Luigi0210 @AlexandervonHumboldt2 @Compassionate @sleepyjess @satellite73 @amistre64 @SolomonZelman

Answer 6

@SolomonZelman @amistre64

Answer 7

@TheSmartOne @Loser66 @Luigi0210 @robtobey @AlexandervonHumboldt2

Answer 8

You have tagged me 5 times on this one post. -.-

Answer 9

I'm sorry ;_;

Answer 10

@Michele_Laino

Answer 11

For both of these you need to use the principle that impulse is the change in momentum.

1. change in momentum = Impulse = force * impact time

2. change in momentum = mass * (final velocity - initial velocity)

Answer 12

@peachpi But couldn't the Impulse-momentum theorem be used? \[F \Delta t = P(final) - P(initial)\]

Answer 13

yes, this is all using the impulse momentum theorem.

Answer 14

So would the change in momentum be 1 N * sec?

Answer 15

yes

Answer 16

awesome! is there a way to figure out how fast it goes afterwards?

Answer 17

Yes, you'd use impulse momentum again.
\[F \Delta t=m \Delta v\]
If you make the assumption that the ball was initially at rest, then solving for \(\Delta v\) gives your final velocity

Answer 18

oh, so would the answer be 2 (N* sec)/kg ?

Answer 19

2 m/s once you do a little work on the units

Answer 20

oh yea, I see what you mean about the units. Thanks for the help, I really appreciate it! You don't mind if I ask you a few more questions on this thread?

Answer 21

no I don't mind

Answer 22

A man (mass man = 50 kg) rushes at 6 m/s to save a 20 kg child standing in front of a moving car. He pushes the child at 7 m/s. What is the man's final velocity after the push? What is his velocity if he grabs the child instead?

Answer 23

I'm a bit confused on what equation to use.

Answer 24

I'm thinking these are collisions
Initial momentum = man's mass * man's initial velocity
Final momentum (push) = man's mass * man's final velocity + child's mass * child's final velocity
Final momentum (carry) = (man's mass + child's mass) * (combined final velocity)

For both situations, momentum is conserved. (initial P = final P)

Push:
(50 kg)(6 m/s) = (50 kg)\(v_f\) + (20 kg)(7 m/s)

Carry
(50 kg)(6 m/s) = (50 kg + 20 kg)\(v_f\)

Answer 25

Oh I see, thank you! I think I switched what equations I used for each situation when I attempted it earlier today. I have another question that I've been struggling with:

8 kids on a "merry go round" are the the outer edge and move to the center. If the original angular velocity was 2 rad/sec and the moment of inertia went from 50 kg * m^2 to 40 kg * m^2, what is the new angular velocity?

Answer 26

hint:
if we can neglect the friction forces, then we can apply the conservation of angular momentum, namely we can apply this formula:
\[{I_1}{\omega _1} = {I_2}{\omega _2}\]
where \(I\) is the momentum of inertia, with respect to the rotation axis. Furthermore, we can write:
\[I = \sum\limits_i {{m_i}r_i^2} \]
and the index \(i\) runs on all the particles which compose the physical system