Collision in two <span title="dimensions!!!!!!!!!!!!!!!?">dimensions!!!!!!!!!!!!!!!...</span>

Question

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

Answer 1

Let

Ma = mass of the first object

Mb = mass of the second object

Va = velocity of the first object

Vb = velocity of the second object

Vf = final velocity of the two objects after the inelastic collision

A = angle of the velocity of mass A (measured from the x-axis)

B = angle of the velocity of mass B (measured from the x-axis)

Equation 1 -- the conservation of momentum along the horizontal axis is

MaVa(cos A) + MbVb(cos B) = (Ma + Mb)Vf

NOTE - it will be assumed that after the collision, the two bodies are moving along the horizontal axis.

Since Ma = Mb, Va = Vb and Vf = 1/2(Va), the above equation is modified to

MaVa(cos A) + MaVa(cos B) = (Ma + Ma)(1/2)Va

MaVa(cos A) + MaVa(cos B) = (2Ma)(1/2)Va

cos A + cos B = 1  --- call this Equation 1

Applying the Law of Conservation of Momentum along the vertical axis,

MaVa(sin A) + MbVb(sin B) = 0

Again, since Ma = Mb and Va = Vb,

then

sin A + sin B = 0  --- call this Equation 2

At this point, there are two equations that have been derived, i.e.,

cos A + cos B = 1  

and

sin A + sin B = 0

Since this is a Physics problem, I trust that you have the required trigonometric skills to proceed and solve for angles A and B.

Good luck.