Inelastic 2D Collision with Vector Components

[JwdePhysics]

Homework Statement

Two balls with mass m and 4m collide at the location x=y=0 and stick. Their initial velocities just before the collision can be represented as v1=(i+j) v and v2=(j-i)v' respectively. Their final velocity vf makes an angle θ with the +x axis. Find v and v' in terms of vf and θ.

Homework Equations

p=mv
Thing x-component = Thing * cos θ (i)
Thing y-component = Thing * sin θ (j)

The Attempt at a Solution

Momentum is conserved meaning that the initial x and y components are equal to the final x and y components of momentum, so:

mv1x-4mv2x=(m+4m)vfcosθ
mv1y+4mv2y=(m+4m)vfsinθ

Simplifying the equations above and removing m yields

v1x-4v2x=5vfcosθ
v1y+4v2y=5vfsinθ

Then we know that the vectors v1 and v2 are equal to (i+j)v and (j-i)v' repsectively. We can represent each vector as

v1 = vi+vj
v2 = -v'i+v'j

where the i represents the x value and j represents the y component.

Now subsituting those components into the equation above yields
vi+4v'i=5vfcosθ
vj+4v'j=5vfcosθ

If I solve for v' first, I get v=(5vfcosθ-4iv')/i if we solve for v in the first equation above, and then subsituting that into the second equation gets j((5vfcosθ-4iv')/i)+4jv'=5vfsinθ

If we distribute the j and multiply the whole thing by i we get
5vfcosθij-4ijv'+4jiv'=5vfisinθ

However the two 4ijv' cancel and then I can't solve for v'. The same thing happens trying to solve for v.
Why does the v'/v keep canceling out?

 

[ehild]

Homework Statement

Two balls with mass m and 4m collide at the location x=y=0 and stick. Their initial velocities just before the collision can be represented as v1=(i+j) v and v2=(j-i)v' respectively. Their final velocity vf makes an angle θ with the +x axis. Find v and v' in terms of vf and θ.

Homework Equations

p=mv
Thing x-component = Thing * cos θ (i)
Thing y-component = Thing * sin θ (j)

The Attempt at a Solution

Momentum is conserved meaning that the initial x and y components are equal to the final x and y components of momentum, so:

mv1x-4mv2x=(m+4m)vfcosθ
mv1y+4mv2y=(m+4m)vfsinθ

Simplifying the equations above and removing m yields

v1x-4v2x=5vfcosθ
v1y+4v2y=5vfsinθ

Then we know that the vectors v1 and v2 are equal to (i+j)v and (j-i)v' repsectively. We can represent each vector as

v1 = vi+vj
v2 = -v'i+v'j

where the i represents the x value and j represents the y component.

Now subsituting those components into the equation above yields
vi+4v'i=5vfcosθ
vj+4v'j=5vfcosθ

i and j are unit vectors along the x and y directions, respectively. You can not substitute components(scalar) with vectors.

 

[JwdePhysics]

i and j are unit vectors along the x and y directions, respectively. You can not substitute components(scalar) with vectors.

Wait what do you mean I can't subsitute vector comonents with vectors? v1x is the x component and v1y is the y component. And i is like the x component of the v1 vector and j is like the y component of the v1 vector

 

[haruspex]

There is a lot of confusion between scalars and vectors here, compounded by the failure to distinguish them typographically.
In the question as stated, v₁, v₂, i and j are vectors, v and v' are scalars. v_f is referred to as a vector, but it will be more convenient to make it a scalar, i.e. the magnitude of the final velocity. That allows you to write v_fcos(θ) etc., which would be problematic if a vector.

mv1x-4mv2x=(m+4m)vfcosθ

Since we need v_f to be scalar, v_1x etc. must be scalars.

vi+4v'i=5vfcosθ

Since the right hand side is scalar, discard the i vectors on the left.

v=(5vfcosθ-4iv')/i

That makes no sense. You cannot divide by a vector.

 

[JwdePhysics]

Since the right hand side is scalar, discard the i vectors on the left.

Oh ok thanks, however if I do discard the i's or j's the v/v' will still cancel out and I'm left with 5vfcosθ=5vfsinθ

 

[JwdePhysics]

Oh so I got it now:
v-4v'=5v_fcosθ and v+v'=5v_fsinθ

and if you solve that system you get v = 5/2v_f(sinθ+cosθ)
and v' = 5/8v_f(sinθ-cosθ)

I took into account the -i when making my equations so there was a +4v' in the first equation instead of a -4v'. Apparently the (j-i) just represents direction and you are only supposed to plug in the scalar v'.[/SUB]

 

[haruspex]

v = 5/2vf(sinθ+cosθ)
and v' = 5/8vf(sinθ-cosθ)

Or, more correctly, v = (5/2)vf(sinθ+cosθ) and v' = (5/8)vf(sinθ-cosθ)

 

[JwdePhysics]

Sorry I came back after a while and noticed I messed up the second equation I wrote v+v'=5vfsinθ when it should be v+4v'=5vfsinθ since the second ball has 4 times the mass.