Momentum in 2 dimensions?

Question

Hey I'm really confused! Please help, I have no idea where to start in even working this out!

A particle A of mass m travelling with a velocity v0 collides with a stationary particle B of mass 4m. After the collision, A travels at right angles to its original direction with speed v1 and B moves with speed v2at an angle of 60° to the original direction of A.

Show that after the collision v2 = [v0/(2sqrt3)] and find an expression for v1

Write down expressions for the total kinetic energies before and after the collision in terms of m and v0. Hence show that the ratio of final to initial kinetic energies is 2/3

Is the collision elastic or inelastic? Explain your answer.

Answer 1

Conservation of linear momentum
http://en.wikipedia.org/wiki/Momentum

the initial momentum must equal the final momentum.

Momentum of a particle equals Mass x Velocity, where velocity is a vector.

Initially, there was one particle moving along a straight line. All the momentum was along that line. Therefore, after the collision, the net momentum (the sum of the momentum vectors of the two particles) must still be on that line.

Particle A ends up with a momentum v1 x m at right angles to the original direction. Therefore particle B must have a component of momentum perpendicular to the original direction which cancels that.

But particle B also is the only one with a velocity component in the original direction, therefore its component in the original direction must carry all of the original momentum or v0 x m.

Since you know the angle that particle B makes with the original direction and you know its component in that direction, you can determine its magnitude and the magnitude of the component in the perpendicular direction.

As for the kinetic energy, Ek = (1/2) m V^2

You know all three velocities in terms of V0 and you know all the masses, so you can compute the relative kinetic energies (i.e. the ratios between them).

Answer 2

You convert the force of the arrow (MassxSpeed) 1000g per 1 kilogram .02 KG (arrow) at 260 M/Sec (speed) = 5.2 Kg M/Sec (Force of Arrow) apply it to the targets weight plus that of the arrow and that will give you the speed of the target (moving in the opposite direction of the arrow.) Be sure to convert to like units of Mass (KG's) 5.2 KG M/Sec (force of arrow) / .820 KG (target weight + Arrow weight) = 6.34 M/Sec The target will maintain the 3.0 M/S of horizontal movement assuming the target was hit exactly perpendicular to its movement. The target will be moving 6.34 M/Sec (N) along with the 3.0 M/S (W) With an arrow the collision must be inelastic, the arrow will not bounce off.