What is the velocity of mass 2m after the collision?

[Xenon]

Homework Statement

Mass of 2m slides on a horizontal plane with the velocity of v0 across the x-axis. And hits the mass m which is in rest. after the collision mass m moves with the velocity of v0/2 in the direction which creates an angle of 30 degrees of the original movmen of 2m.
a) What is the velocity of mass 2m after the collision?
b) What is the direction of mass 2m after the collision?
c) Is the collision described is perfectly ellastic collision?

Homework Equations

P=m*v
Pi=Pf

The Attempt at a Solution

I have broken down the velocity to x,yPinitialx=Pfinalx
m1*V1ix+m2*V2ix=m1*V1fx+m2*v2fx
2m*v0+m*0 = 2m*Vfcosθ+m*v0/2*cos30Pintialy=Pfinaly
m1*V1iy+m2*V2iy=m1*V1fy+m2*v2fy
0 = -2m*Vfsinθ+m*V0/2sin30

I don't know what to do from here. i can't simplfy the equations,
could you help me with that?

Thank you,

 

[rolotomassi]

You have 2 unknowns so you want to get two expressions that arise from conservation of momentum in both x and y directions. You will have two simultaneous equations which you can solve for the unknowns. I am not sure if you have already done that its hard to follow those expressions but once you get your two equations its relatively straight forward to solve by substitution. Remember that your answer will be in terms of $$m\ and \ v_0$$

 

[insightful]

0 = -2m*Vfsinθ+m*V0/2sin30

I think you meant "V0/2*sin30" as in (V0/2)*sin30.

Hint: sinx/cosx = tanx

 

[Xenon]

My problem is that i just can't simplfy it. could you show me what to do with the parameters? V0 and the θ angles.
Thank you for helping me.

 

[SammyS]

My problem is that i just can't simplfy it. could you show me what to do with the parameters? V0 and the θ angles.
Thank you for helping me.

Your figure has V_f as u. That's a bit handier.

What are the values of sin30° and cos30° ?

Solve 0 = -2m*u*sinθ+m*(u/2)sin30 for u. (EDIT: This is corrected below.)

Solve 0 = -2m*u*sinθ+m*(V₀/2)sin30 for u.​

Plug that result into the other equation:

2m*u+m*0 = 2m*u*cosθ+m*(u/2)*cos30

2m*u+m*0 = 2m*u*cosθ+m*(V₀/2)*cos30

2m*V₀+m*0 = 2m*u*cosθ+m*(V₀/2)*cos30​

Solve for cotθ (Lots of stuff cancels.)

 

[Xenon]

U=-m*sin tetha-sin30/2 ?

 

[insightful]

Solve 0 = -2m*u*sinθ+m*(u/2)sin30 for u.

Doesn't u cancel from this equation?

 

[SammyS]

Doesn't u cancel from this equation?

DUH! Yes it does!

Thanks for noticing.

I had a major typo there!

It's now been corrected. (I hope correctly!)

Solve 0 = -2m*u*sinθ+m*(V₀/2)sin30 for u.

 

[Xenon]

why does it cancels?

umm,
iam really confused why does it cancels?
we are left with u/2 don't we?

 

[insightful]

2m*u+m*0 = 2m*u*cosθ+m*(V0/2)*cos30

Shouldn't the first u in this equation be V₀?

 

[SammyS]

Shouldn't the first u in this equation be V₀?

Yup !

 

[Xenon]

This is what i got:
0 = -2m*u*sinθ+m*(V0/2)sin30 for u.
u=-m*sinθ+0.5v0*sin30
2m*V0+m*0 = 2m*u*cosθ+m*(V0/2)*cos30 solve for cotθ
cosθ=u+0.5v0*30

when plugging u into cosθ to solve for cotθ

cotθ=-m+0.5v0*sin30+0.5v0*cos30

What now?
Thank you for all of your help :)

 

[SammyS]

This is what i got:
0 = -2m*u*sinθ+m*(V0/2)sin30 for u.
u=-m*sinθ+0.5v0*sin30
2m*V0+m*0 = 2m*u*cosθ+m*(V0/2)*cos30 solve for cotθ
cosθ=u+0.5v0*30

when plugging u into cosθ to solve for cotθ

cotθ=-m+0.5v0*sin30+0.5v0*cos30

What now?
Thank you for all of your help :)

The mass, m, cancels out of both equations.

You did not solve for u correctly in the first equation.

Your result for the second equation is also far from correct.

 

[Xenon]

Could you please show me how to do it correctly,
I tried to do it, bit by bit... and it still not right.

Thanks

 

[SammyS]

Could you please show me how to do it correctly,
I tried to do it, bit by bit... and it still not right.

Thanks

I'll get you started.

To solve 0 = -2m⋅u⋅sinθ+m⋅(V₀/2)sin30 for u, subtract the first term from both sides. (Add its opposite.)

2m⋅u⋅sinθ = m⋅(V₀/2)sin30

Then divide both sides by 2m⋅sinθ. Right? What do you get?

 

[Xenon]

I got
u=05V0*sin30/2sinθ

 

[SammyS]

I got
u=05V0*sin30/2sinθ

Using the "reply" feature will help you make more readable posts. Then you can cut & paste more accurately. Also, to make a subscript, use the X₂ icon on the green strip above the message box you use to form your message to post.

You should use parentheses around that denominator. The leading 05 is missing a decimal point. You might as well divide 0.5 by 2 .

u = (0.5)V₀⋅sin30/(2⋅sinθ)

u = (0.25)V₀⋅sin30/(sinθ)​

By the way, do you know the value of sin(30°) ?

 

[Xenon]

So
u = (0.125)V0/(sinθ)
How is that looking?

Thank you very much, you have no idea how glad iam, that you are helping me.