What is the change of momentum of the ball?

[underduck]

1. A 0.095 kg tennis ball is traveling 40 m/s, hits a wall and travels in the opposite site direction it came from. The bounced leaving the wall with a speed of 30 m/s.

a. What is the change of momentum of the ball?

b. If the time contact is 20ms, find the impulse force.

m1= 0.095 v1= 40 m/s
m2= 0.095 v2= -30 m/s
My equation for:

a. change of momentum= (m1 x v1) - (m2 x v2)
(0.095 x 40) - (0.095 x -30) = 6.65 kg m/s
is this equation right?

b. i have no idea. i need some help here please :)

 

[CWatters]

a) A change in momentum would be = momentum after - momentum before. It's also a vector quantity so is the ball going in the same direction afterwards?

b) Impulse = Change in momentum = force * time

 

[Redbelly98]

a. change of momentum= (m1 x v1) - (m2 x v2)
(0.095 x 40) - (0.095 x -30) = 6.65 kg m/s
is this equation right?

Yes, correct.

For (b), see the post by CWatters.

p.s. Welcome to Physics Forums.

 

[underduck]

Yes, correct.

For (b), see the post by CWatters.

p.s. Welcome to Physics Forums.

Answer for

B. F= (mv - mu)/t
= (0.095 x -30) - (0.095 x 40) / 20
= -0.33 N

is that correct ?
hehe my first time doing physics. thanks for the help guys!

 

[CWatters]

I dissagree with the answer you give for a)

The change in momentum = MV₂ - MV₁

m1= 0.095 v1= 40 m/s
m2= 0.095 v2= -30 m/s

Therefore the change

= (0.095 * -30) - (0.095 * 40) = -2.85 - 3.8 = -6.65 kg.s^-1

Note the minus sign.

 

[CWatters]

For b)

The impact force = change of momentum / time (in seconds)

= -6.65/(20 x 10^-3)
= -332.5 N

Negative because the applied force is in the opposite direction to the initial direction the ball was traveling in.

 

[underduck]

I dissagree with the answer you give for a)

The change in momentum = MV₂ - MV₁

m1= 0.095 v1= 40 m/s
m2= 0.095 v2= -30 m/s

Therefore the change

= (0.095 * -30) - (0.095 * 40) = -2.85 - 3.8 = -6.65 kg.s^-1

Note the minus sign.

For b)

The impact force = change of momentum / time (in seconds)

= -6.65/(20 x 10^-3)
= -332.5 N

Negative because the applied force is in the opposite direction to the initial direction the ball was traveling in.

I got it for the first question.

for the second one there is one thing I'm confused there:
(20 x 10^-3)

why do i need to add 10^-3 ?

thank you. i really appreciate your help, making things easier for me.

 

[underduck]

I have another question on this Impulse and Momentum topic.

Two identical blocks, each mass 10kg, are used in an experiment. The First is held at rest on 20° incline plane 10m from the second, which at rest at the foot of the plane. the one descends the incline, slams into and sticks to the second, and the sail off together horizontally. given μk= 0.35, g 9.8 m/s2


1. Find the velocity of the first block just before it hit the second block

2. Find the velocity after the impact

3. Find the change of kinetic energy

 

[CWatters]

why do i need to add 10^-3 ?

The contact time is 20ms not 20s.

20ms = 20 * 10^-3seconds.

 

[CWatters]

I have another question on this Impulse and Momentum topic.

Which bit are you stuck on? Have you drawn a diagram showing the forces on the first block?

 

[underduck]

Which bit are you stuck on? Have you drawn a diagram showing the forces on the first block?

I'm stuck from the first question. getting a lil bit confused on how to use the equation. maybe you can explain.

Here i attach the free body diagram for the question. thank you

 

[Redbelly98]

There are problems with your free body diagram:

1. There is no applied force Fapp. (There would have been a force holding the block in place initially, but then it is released and allowed to slide down the incline.)

2. There are other forces involved that you did not draw for the top block. The "μ_k" and "g" suggest what two of those forces are.