What Is the Angle Between Two Objects' Velocities After an Inelastic Collision?

[mb85]

After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/4 their initial speed. Find the angle between the initial velocities of the objects.

I figured.. 2m(v/4)
but then i just get lost... can someone help! thanks!

 

[nrqed]

After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/4 their initial speed. Find the angle between the initial velocities of the objects.

I figured.. 2m(v/4)
but then i just get lost... can someone help! thanks!

Pick a direction for the first object, let's say East. Let's say the second object is coming from an angle \theta North of East. Then Decompose their momenta into x and y components, apply conservation of momentum along x and y directions. You will find the final x and y velocities along x and y in terms of the initial speed v and the angle \theta. Impose that {\sqrt { v_{x final}^2 + v_{y final}^2}} = v/4 and that will give you a single equation for \theta.


Patrick

 

[BerryBoy]

OK, so the collision is in 2 dimensions, so you know you're going to be using vectors. So let me start you off with what you know.

m(\underline{v}_1 + \underline{v}_2 ) = 2m \underline{v}_{final}

Do you know how to make vectors loose their directional components?

(Hint: Multiply both sides by a vector you know)

Hope this Helps, Sam

 

[BerryBoy]

OK, pick the vector carefully.

I have just done this question in 5 lines (instead of 10) by picking another vector. If you don't choose your vector well, you'll have to use some trig. identities (which I don't like if I can avoid it).

Draw a diagram and note what the angles have in relation to each other due to symmetry. If you want, I'll put up a diagram of what I'm trying to say... just ask if you want it.

Regards,
Sam

 

[mb85]

hey thanks. this is what i did so far. but I am still have trouble.

M1V1i + M2V2i = M1V1f + M2V2f

initial for 1
X = mv cos Theta
y = mv sin theta

final for 1
X = mv
y=0

Initial for 2
X = mv cos theta
Y = mv sin theta

final for 2
X =mv
y = 0


mv cos theta + mv sin theta = MVi
Mv (cos theta + sin theta) = MVi

? then i get lost.

 

[vaishakh]

The components of their initial velocities perpendicular to the direction of their final velocity adds to zero. And the components along this direction must thus be same and add upto v/4.

 

[vaishakh]

Sorry. The second sentence is - the component of their initial velocities along the direction of final velocity must thus be same and add upto v/4

 

[mb85]

im still confused.

can someone show me the steps?

 

[vaishakh]

Let a be the angle made with their initial velocities along the direction of final veloctitis.
Now,
v*cosa + v*cosa = v/4.
Thus cosa = 1/8