Conservation of momentum and mechanical energy on an inclined plane

[Spector989]

Homework Statement

A particle of mass m slides down the smooth inclined face of a wedge of mass 2m, and inclination a, which is free to move on a smooth horizontal table. Use equations of momentum and energy to obtain an expression for the velocity of the particle relative to the wedge when has moved a relative distance s from rest down the inclined face of the wedge.

Relevant Equations

Pi =Pf , P.Ei + K.E i = P.Ef + K.Ef

So i am tried to conserve momentum and use conservation of mechanical energy but won't there be psuedo force acting on the block if i am solving from non inertial frame ?. If i ignore the pseudo force and simply use C.O.M.E and include the K.E of the wedge and solve normally i do get the appropriate answer but that doesn't fit right to me .

 

[Lnewqban]

Welcome @Spector989 !

Could you show us the work that led you to get the appropriate answer?

Why not to solve the problem using an inertial frame of reference, considering the relative movements of particle and wedge?
It seems to be the easiest approach to me.

 

[haruspex]

Why not to solve the problem using an inertial frame of reference

That is not relevant to the question being asked by the OP, namely, how come the non-inertial frame gave the official answer. But yes, we need to see the actual working to answer that.

 

[Spector989]

Welcome @Spector989 !

Could you show us the work that led you to get the appropriate answer?

Why not to solve the problem using an inertial frame of reference, considering the relative movements of particle and wedge?
It seems to be the easiest approach to me.

I rechecked what i did and i solved it wrong .. but i did what you told me to and i can't understand what to do further . Btw i took the intial position of block as reference line for gravitational P.E

 

[Spector989]

That is not relevant to the question being asked by the OP, namely, how come the non-inertial frame gave the official answer. But yes, we need to see the actual working to answer that.

Mb i solved it wrong , but i tried it solving using inertial frame and i am stuck , will the final velocity of the wedge be different, will the normal reaction between wedge and block be considered internal force if i take them as a system but only objects with same acceleration can be taken as a system right ?

 

[haruspex]

Mb i solved it wrong , but i tried it solving using inertial frame and i am stuck , will the final velocity of the wedge be different, will the normal reaction between wedge and block be considered internal force if i take them as a system but only objects with same acceleration can be taken as a system right ?

Please, please post the working that you did using a non inertial frame to arrive at the correct answer. Preferably typed in, not as an image. Images are for textbook extracts and diagrams. If you must do it as an image, please make sure it is easy to read and laid out in a sequence, not scattered all over the page.

In the working you posted, it is unclear what you mean by velocity of block. In which direction?

 

[Spector989]

Please, please post the working that you did using a non inertial frame to arrive at the correct answer. Preferably typed in, not as an image. Images are for textbook extracts and diagrams. If you must do it as an image, please make sure it is easy to read and laid out in a sequence, not scattered all over the page.

In the working you posted, it is unclear what you mean by velocity of block. In which direction?

Sorry if i was not clear , The workings i did to arrive at the correct answer through non inertial frame was Wrong , and i have mentioned the direction in the right corner beside the FBD for the block and on the left side of the page for the wedge . I cannot type in the answer as i am not used to typing out equations and i am on a mobile . I solved it by trying it again through inertial frame of reference. I can post it but i can't type it out. I got a relation between Velocity of wedge and block but the velocity of the block will not be constant so i am getting different variables in both sides of the eqn while applying conservation of mechanical energy

 

[PeroK]

Sorry if i was not clear , The workings i did to arrive at the correct answer through non inertial frame was Wrong , and i have mentioned the direction in the right corner beside the FBD for the block and on the left side of the page for the wedge . I cannot type in the answer as i am not used to typing out equations and i am on a mobile . I solved it by trying it again through inertial frame of reference. I can post it but i can't type it out. I got a relation between Velocity of wedge and block but the velocity of the block will not be constant so i am getting different variables in both sides of the eqn while applying conservation of mechanical energy

I think your equation ##v = \frac{u\cos \alpha}{3}## is correct.

Can you figure out how to calculate ##v## in terms of ##s## using conservation of energy?

 

[Spector989]

I think your equation ##v = \frac{u\cos \alpha}{3}## is correct.

Can you figure out how to calculate ##v## in terms of ##s## using conservation of energy?

That is the part i am having trouble in , can you help me a bit through it.

 

[PeroK]

That is the part i am having trouble in , can you help me a bit through it.

I would work in the IRF. You need to relate the KE of the mass and the wedge to the loss of PE. You need a diagram that shows the mass having slid a distance ##s## down the wedge and the new position of the wedge and the mass. You should have a triangle with base of length ##x + X##, height ##y## and hypoteneuse ##s##, where ##x## is the horizontal distance moved by the mass, ##X## is the horizontal distance moved by the wedge and ##y## is the vertical distance the mass has dropped.

 

[Spector989]

I would work in the IRF. You need to relate the KE of the mass and the wedge to the loss of PE. You need a diagram that shows the mass having slid a distance ##s## down the wedge and the new position of the wedge and the mass. You should have a triangle with base of length ##x + X##, height ##y## and hypoteneuse ##s##, where ##x## is the horizontal distance moved by the mass, ##X## is the horizontal distance moved by the wedge and ##y## is the vertical distance the mass has dropped.

Thanks a lot , i took the top of the triangle as reference line and conserved momentum and i got the required answer . Thanks:)

 

[Lnewqban]

Thanks a lot , i took the top of the triangle as reference line and conserved momentum and i got the required answer . Thanks:)

Could you show us your final work?

 

[Spector989]

Could you show us your final work?

 

[PeroK]

Thanks a lot , i took the top of the triangle as reference line and conserved momentum and i got the required answer . Thanks:)

I did it slightly differently. Starting from:

A triangle with base of length ##x + X##, height ##y## and hypoteneuse ##s##, where ##x## is the horizontal distance moved by the mass, ##X## is the horizontal distance moved by the wedge and ##y## is the vertical distance the mass has dropped.

We have:
$$x + X = s \cos \alpha \ \Rightarrow \ v_x + V = u\cos \alpha$$$$mv_x = 2mV \ \Rightarrow \ v_x = \frac 2 3 u\cos \alpha, \ V = \frac 1 3 u\cos \alpha$$$$y = s \sin \alpha \ \Rightarrow \ v_y = u \sin \alpha$$By conservation of energy:
$$mgy = \frac 1 2 m(v_x^2 + v_y^2) + \frac 1 2 (2m)V^2$$$$\Rightarrow 2gs \sin \alpha = \big (\frac 4 9 \cos^2 \alpha + \sin^2 \alpha + \frac 2 9 \cos^2 \alpha \big ) u^2$$$$\Rightarrow u^2 = \frac{6gs\sin \alpha}{2 + \sin^2 \alpha}$$

 

[Spector989]

I did it slightly differently. Starting from:We have:
$$x + X = s \cos \alpha \ \Rightarrow \ v_x + V = u\cos \alpha$$$$mv_x = 2mV \ \Rightarrow \ v_x = \frac 2 3 u\cos \alpha, \ V = \frac 1 3 u\cos \alpha$$$$y = s \sin \alpha \ \Rightarrow \ v_y = u \sin \alpha$$By conservation of energy:
$$mgy = \frac 1 2 m(v_x^2 + v_y^2) + \frac 1 2 (2m)V^2$$$$\Rightarrow 2gs \sin \alpha = \big (\frac 4 9 \cos^2 \alpha + \sin^2 \alpha + \frac 2 9 \cos^2 \alpha \big ) u^2$$$$\Rightarrow u^2 = \frac{6gs\sin \alpha}{2 + \sin^2 \alpha}$$

This is the first time i have ever seen someone solve a wedge and block system by isolating the distance travelled and making a seperate triangle. Honestly makes the process a lot more cleaner and easier :)

 

[haruspex]

This is the first time i have ever seen someone solve a wedge and block system by isolating the distance travelled and making a seperate triangle. Honestly makes the process a lot more cleaner and easier :)

The question asked in post #1 was how come a solution using a non inertial frame but leaving out the pseudo force got the right answer. For whatever reason, the OP refuses to post that working, so it is hard to answer that question. But it may well be that the OP effectively used the method in post #14, not really making use of a non inertial frame at all.

 

[Spector989]

The question asked in post #1 was how come a solution using a non inertial frame but leaving out the pseudo force got the right answer. For whatever reason, the OP refuses to post that working, so it is hard to answer that question. But it may well be that the OP effectively used the method in post #14, not really making use of a non inertial frame at all.

I mentioned it earlier too , the workings i did for getting the right answer through non inertial frame was wrong .

 

[Spector989]

How do i close a thread?

 

[kuruman]

How do i close a thread?

You don't. Only mentors have the superpowers to do that.

 

[haruspex]

I mentioned it earlier too , the workings i did for getting the right answer through non inertial frame was wrong .

Yes, I saw that, but if it got the right answer despite being wrong then I would like to see those wrong workings to figure out how that can be.