Inelastic collision. Two angle variables.

[ttttrigg3r]

http://session.masteringphysics.com/problemAsset/1007938/22/6318.jpg
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle phi south of east (as indicated in the figure). After the collision, the two-car system travels at speed v_final at an angle theta east of north.

Now I solved Vfinal into its i and j components

i: v_icos[itex]\phi[/itex]=2v_fcos[itex]\theta[/itex]

j: 2v_i-sin[itex]\phi[/itex]v_i=2(v_fsin[itex]\theta[/itex])

The problem is if I solve for vfinal, I get the answer in terms of Vinitial, phi, and theta. The problem only asks for Vfinal in Vinitial and theta. How do I get rid of phi to only have theta and Vi?

 

[Matapurga]

I think you changed the equations.It should be
i: vicosϕ=2vfsinθ
j: 2vi-sinϕvi=2(vfcosθ)
And my sugestion to get rid of phi is to isolate cosϕ on eq.i and use the fundamental relation cos²ϕ+sin²ϕ=1 to find sinϕ, and then substitute sinϕ on eq.j.

 

[ttttrigg3r]

do you mean isolate cos(phi) and square it, and then isolate sin(phi) and then square it and then sub it into cos^2phi + sin^2phi =1? I did that but then I got an equation Vf and Vf^2 . The fraction looks unsolvable for Vf. Cannot factor or anything.

 

[quietrain]

if i am not wrong

you can rearrange it to

2v_i - 2v_fcos(t) = vsin(phi)

2v_fsin(t) = vcos(phi)

square both sides

your right hand side will be v_i² , since c and s cancel

your left hand side will be

4v_i² + 4v_f² - 8v_iv_fcos(t)

this is quadratic, so use -b+- squareroot b^2 - 4ac over 2a.

 

[ttttrigg3r]

if i am not wrong

you can rearrange it to

2v_i - 2v_fcos(t) = vsin(phi)

2v_fsin(t) = vcos(phi)

square both sides

your right hand side will be v_i² , since c and s cancel

your left hand side will be

4v_i² + 4v_f² - 8v_iv_fcos(t)

this is quadratic, so use -b+- squareroot b^2 - 4ac over 2a.

I don't see how the sin and cos cancel when you square it? There must be a step I'm missing because if you square the first equation, on the right side you get Vi²sin[itex]\phi[/itex]. On the left side you would get a long quadratic with variables Vi, Vf, and theta.

 

[quietrain]

I don't see how the sin and cos cancel when you square it? There must be a step I'm missing because if you square the first equation, on the right side you get Vi²sin[itex]\phi[/itex]. On the left side you would get a long quadratic with variables Vi, Vf, and theta.

the aim is to get rid of the phi term right?

so that means you have to express phi in terms of everything else

so squaring both sides of both equations give you

4v_i² - 8v_iv_fcos(t) + 4v_f²cos²(t) = v_i²sin²(phi)

4v_f²sin²(t) = v_i²cos²(phi)

so now look at the RHS of both equations, you add them so that the cos and sin (phi) "cancel off" since cos²x + sin²x = 1, so you effectively left with v_i² , i.e, you have factored out v_i²

so similarly, for lhs, you combine the sin and cos (t) to get

4v_i² + 4v_f² - 8v_iv_fcos(t) = v_i²

so you are now effectively solving for v_f in a quadratic equation, without the phi terms