What are the applications of the Conserved Momentum Theorem?

[MasterNe0]

Hi,
I need help with these 3 conserved momentum theorem using the formula:
(Pa2) + (Pb2) = (Pa1) + (Pb1)

1) A 2575 kg van runs into the back of a 825 kg compact car at rest. They move off together at 8.5 m/s. Assuming the friction with the road can be negligible,find the initial speed of the van.

2) A 0.115 kg hockey puck moving at 35.0 m/s strikes a 0.265 kg octopus thrown onto the ice by a hockey fan. The puck and octopus slide off together. Find their velocity.

3) A 50 kg woman riding on a 10 kg cart is moving east at 5.0. The woman jumps off the front of the cart and hits the ground at 7.0 eastward. Relative to the ground. Find the velocity of the cart after the woman jumps off.



Any help in solving theses 3 will be appreciated.

 

[Pyrrhus]

Just use

$$\Delta \vec{p} = 0$$

Momentum is conserved for example on the first problem

$$\vec{p}_{van} + \vec{p}_{car} = \vec{p}_{van+car}$$

 

[Banana]

These are inelastic collisions, meaning after the 2 things hit, they stick together. So you can treat the 'after' part of the equation as one thing. That's what Cyclovenom is saying. The next step is to replace momentum with mass and velocity: m1 * v1 = (m1 + m2)v'.
Note: v' stands for velocity of the blob after the wreck. For the first problem, you're looking for v1 because you let the thing that is initially moving be Object 1.
Step 2 is to rearrange the equation to get the thing you're looking for isolated. When you fill the rest in, it is best to take SI units into the equation to be sure everything comes out as expected.

 

[Raza]

Question:
A 2575 kg van runs into the back of a 825 kg compact car at rest. They move off together at 8.5 m/s. Assuming the friction with the road can be negligible,find the initial speed of the van.

Solution:(Note that this MIGHT be the right solution. I made it up)
2575kg+825kg=3400kg
So the mass of 3400kg moves at 8.5m/s and you are trying to find the velocity when it is at 2575kg
$$\frac{8.5m/s}{3400kg} = \frac{x}{2575kg}$$
In this case, you divide 8.5m/s by 2575kg and then multiply that with 3400kg. When you do that, you get 11.2m/s as your velocity.

Again, I might be wrong.
*Edit: I was right

 

[MasterNe0]

i was thinking along the line of that answer when i was trying to do it. I am not sure thought so not sure if it correct.

Any help with the other 2 problems?

 

[futb0l]

2) A 0.115 kg hockey puck moving at 35.0 m/s strikes a 0.265 kg octopus thrown onto the ice by a hockey fan. The puck and octopus slide off together. Find their velocity.

Just using the formula and simple substitution...

$$m_1v_1 + m_2v_2 = (m_1+m_2)v'$$

$$0.115(35) = (0.25+0.115)v'$$

3) A 50 kg woman riding on a 10 kg cart is moving east at 5.0. The woman jumps off the front of the cart and hits the ground at 7.0 eastward. Relative to the ground. Find the velocity of the cart after the woman jumps off.

same stuff here... just calculate the momentum of before and after... and they should be equal to each other.

$$(m_{woman}+m_{cart})v = (m_{woman}*7) + m_{cart}v'$$

 

[Raza]

Question:
A 0.115 kg hockey puck moving at 35.0 m/s strikes a 0.265 kg octopus thrown onto the ice by a hockey fan. The puck and octopus slide off together. Find their velocity.

Solution:(Note that this MIGHT be the right solution. I made it up)
0.115kg+0.265kg=0.38kg
So the mass of 0.0.115kg moves at 35.0m/s and you are trying to find the velocity when it is at 0.38kg
$$\frac{35m/s}{0.115kg} = \frac{x}{0.38kg}$$
In this case, you divide 35m/s by 0.38kg and then multiply that with 0.115kg. When you do that, you get 10.59m/s as your velocity.

I tested it to see if I was right by also using the equation in futb0l's post and got the same answer.

 

[MasterNe0]

Ok. thanks a lot.