Work,energy COM and linear momentum conservation

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Homework Statement

A wagon with mass M can move on frictionless surface. A mathematical/ideal pendulum is fastened on the wagon. At the initial moment the wagon and the pendulum were at rest and the pendulum makes an angle of x with the vertical. What will be the velocity of the wagon when the pendulum makes an angle of y with the vertical?

Homework Equations

I guess its a linear momentum question but I don't know the equation.

The Attempt at a Solution

 

[hikaru1221]

Hi,

1 - You nearly got the 1st thing right. Since friction is neglected, the system wagon-pendulum experiences no horizontal force, and thus its horizontal component of linear momentum is conserved. From this, you have the 1st equation relating the horizontal components of the velocities of the wagon and the bob $$v_{h(wagon)}$$ and $$v_{h(bob)}$$ respectively.

2 - The 2nd conserved thing is the total energy of the system, which comprises of the potential energy of the bob and the kinetic energies of the wagon and the bob. This is the 2nd equation, which relates $$v_{h(wagon)}$$ , $$v_{h(bob)}$$ , $$v_{v(bob)}$$ - the vertical component of the bob's velocity and the angles x, y.

3 - The 3rd condition is that in the reference frame of the wagon, the velocity of the bob is perpendicular to the string. You should deduce the 3rd equation relating $$v_{h(wagon)}$$ , $$v_{h(bob)}$$ , $$v_{v(bob)}$$ and the angle y.

Try to write down the equations

 

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well i m not sure of the first equation.
is it mv(bob)= - Mv(wagon)??

 

[hikaru1221]

It's $$mv_{h(bob)} = - Mv_{h(wagon)}$$. (horizontal component!)

 

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but the velocity of bob is it relative to the ground??

 

[hikaru1221]

Yes. If you choose the reference frame of the wagon from the start, you have to take into account the fictitious force, which is unnecessarily complicated.