Not sure where to start(momentum problem, I think)

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The discussion focuses on solving two momentum problems involving a wagon and an explosion. In the first scenario, two children jump off a 150kg wagon, with the 40kg boy jumping at 4m/s and the 50kg girl at 2m/s. By applying the conservation of linear momentum, the wagon's speed is calculated to be 2.5m/s in the same direction as the girl. The second problem involves an explosion that splits a rock into three pieces, requiring the use of momentum conservation in both x and y components to determine the mass and direction of the third piece moving at 30m/s.

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Please help me with this, as I'm not even sure where to start.

Two children are standing on a 150kg wagon which is at rest. The children jump off the wagon in opposite directions. The 40kg boy jumps off at 4m/s and the 50kg girl jumps off at 2 m/s. What is the speed and direction of the wagon?

As well as this one

An explosion blows a rock into three parts. Two pieces fly off at right angles to one another. A 1.0Kg piece moves at 12m/s and a 3kg piece moves at 8m/s. Determine the mass and direction of motion of the third piece if it has a speed of 30m/s.
 
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Use conservation of linear momentum in both cases.
 
You are right, both of these questions are infact momentum questions. In either question, momentum must be conserved. For question 1, what is the momentum of the girl and the boy? What then must be the momentum of the wagon?

For question two, again, conservation of momentum. However, this time it gets a little harder. You have to break the momentum up into components and then solve it.
 
Lets divide the problem in two parts: initial and final.

INITIAL:

"Two children are standing on a 150kg wagon which is at rest."

So, the initial momentum is 0 because the entire system is at rest.

FINAL:

"The 40kg boy jumps off at 4m/s and the 50kg girl jumps off at 2 m/s"

Lets assume that the boy jumps from the left, and the girl from the right of the wagon, so the momentum of the boy is in -x direction and the girl's is in x direction. The wagon? We don't know yet.


The final momentum is \vec{p}_{boy} + \vec{p}_{girl} + \vec{p}_{wagon} = -40 \cdot 4 \vec{e_x} + 50 \cdot 2 \vec{e_x} + 150 \cdot \vec{v_w}

The conservation of momentum says that \vec{p_{initial}} = \vec{p_{final}}, so

0 = -40 \cdot 4 + 50 \cdot 2 + 150 \cdot v_w \Rightarrow 60 = 150 \cdot v_w \Rightarrow v_w = 2.5 m/s

And as v is positive, the wagon goes in the same direction the girl does.

The other problem is the same.
 
Last edited:
6/15=2.5?++
I think NOT.
 

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