Cart-Cart Collision, Conservation of Momentum

[pleasehelpme6]

Homework Statement

A cart (m1 = 130 kg) is moving to the right along a track at v1i = 24 m/s when it hits a stationary cart (m2 = 350 kg) and rebounds with a speed of v1f = 4 m/s in the opposite direction.

a) With what speed does the 350 kg cart move after the collision?
A: 10.4 m/s

An observer moves in the same direction as the incoming cart with a speed of 12 m/s.
Using the convention that the positive direction is to the right, what are the following velocities with respect to this observer:

b) v1i, ob... A: 12
c) v2i, ob... A: -12
d) v1f, ob... A: 16
e) v2f, ob... A: -1.6

This is the annoying part...

f) What is the total momentum of the system before the collision as seen by this moving observer?

g) What is the total momentum of the system after the collision as seen by this same observer?

Homework Equations

m1v1 = m2v2

Any thoughts on how to solve this? I'm a little lost.

 

[Axiom17]

Think about 1. the definition of momentum, and 2. conservation of momentum.

 

[pleasehelpme6]

yeah i tried using conservation of momentum but the answers I am getting don't make sense to me. the equation doesn't even out.

the way i think of it, from a stand-still perspective, the net momentum is 0, so if youre moving at 12 m/s, shouldn't the net momentum look like -12 m/s?

i tried using the new velocities in the m1v1 = m2v2 equation, but i think i must be setting it up wrong somehow. any suggestions?

 

[Axiom17]

Perhaps this diagram http://yfrog.com/45pf2cj will help somewhat (sorry it's somewhat crudely been done in MS Paint)

Surely the momentum should always be the same, regardless of whether the observer is moving or not, right?!

If you take the definition:

[tex]m_{1}v_{1i}+m_{2}v_{2i} = m_{1}v_{2i}+m_{2}v_{2f}[/itex]

Using all the values calculated in the first part, you should find that both sides of the equation are equal, hence momentum is conserved.

Now if you also alternatively use the values calculated in respect to the observer, you should find that again both sides of the equation are equal and hence momentum conserved.

Try doing those, and post you're calculations if you still have any problems.

 

[rdl]

To solve this problem, we can use the conservation of momentum principle, which states that the total momentum of a system remains constant before and after a collision. In this case, the system includes both carts and the observer.

a) According to the conservation of momentum principle, we can set the initial momentum of the system equal to the final momentum of the system. This can be written as:

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the given values, we get:

130 kg * 24 m/s + 350 kg * 0 m/s = 130 kg * 4 m/s + 350 kg * v2f

Solving for v2f, we get:

v2f = 10.4 m/s

b) According to the observer's frame of reference, the initial velocity of the incoming cart is 12 m/s, since the observer is moving in the same direction as the cart. Therefore, v1i, ob = 12 m/s.

c) The initial velocity of the stationary cart is 0 m/s in the observer's frame of reference, since the cart is not moving relative to the observer. Therefore, v2i, ob = 0 m/s.

d) After the collision, the incoming cart has a velocity of 4 m/s in the opposite direction, but the observer is still moving at 12 m/s. Therefore, the final velocity of the incoming cart in the observer's frame of reference is v1f, ob = -12 m/s + 4 m/s = 16 m/s.

e) After the collision, the stationary cart has a velocity of 10.4 m/s in the opposite direction. But since the observer is moving in the same direction as the incoming cart, the final velocity of the stationary cart in the observer's frame of reference is v2f, ob = 12 m/s - 10.4 m/s = 1.6 m/s.

f) The total momentum of the system before the collision can be calculated by adding the individual momentums of the carts and the observer. In this case, it would be:

m1v1i + m2v2i + mv, ob = 130 kg * 24 m/s + 350 kg * 0 m/s + 12 kg * 12 m/s = 3120 kg