Perfectly Elastic Collision Arrow / Finding Final Velocity

[De_Dre01]

Homework Statement

[/B]
Person A fires a 222 g arrow towards an archery target at a speed of 109 m/s. Person B shoots a 190. g arrow moving in the same direction. This arrow moves with a speed of 290. m/s, catches up, and then collides with Person A's arrow.

If the arrows collide in a perfectly elastic manner, what will be the magnitude of velocity of Person B's arrow afterwards?

Homework Equations

Conversation of Momentum
m1v1+m2v2=m1v1'+m2v2'

The Attempt at a Solution

So far, I tried doing:

m1v1+m2v2=m(1+2)vf'
(0.222)(109)+(0.190)(290)=(0.222+0.190)vf'
And solved for vf. but it isn't the correct answer.

I also tried Conversation of Kinetic Energy
1/2(0.222)(109)^2+1/2(0.19)(290)^2=1/2(0.412)v^2
And solved for vf. but it isn't the correct answer.

What am I doing wrong?

 

[SammyS]

Homework Statement

[/B]
Person A fires a 222 g arrow towards an archery target at a speed of 109 m/s. Person B shoots a 190. g arrow moving in the same direction. This arrow moves with a speed of 290. m/s, catches up, and then collides with Person A's arrow.

If the arrows collide in a perfectly elastic manner, what will be the magnitude of velocity of Person B's arrow afterwards?

Homework Equations

Conversation of Momentum
m1v1+m2v2=m1v1'+m2v2'

The Attempt at a Solution

So far, I tried doing:

m1v1+m2v2=m(1+2)vf'
(0.222)(109)+(0.190)(290)=(0.222+0.190)vf'
And solved for vf. but it isn't the correct answer.

Well, the above is for a perfectly inelastic collision.

I also tried Conversation of Kinetic Energy
1/2(0.222)(109)^2+1/2(0.19)(290)^2=1/2(0.412)v^2
And solved for vf. but it isn't the correct answer.

What am I doing wrong?

The arrows have different velocities after the collision as well as before.

 

[gneill]

Your relevant equation is pertinent. It describes the conservation of momentum for an elastic collision.

What you wrote in your first attempt was an equation for a perfectly inelastic collision (where the projectiles stick together upon collision). This does not fit the described scenario which is an elastic collision, so it led to an incorrect result.

For your second attempt you tried conservation of kinetic energy, which does in fact hold for an elastic collision, but once again you assumed the collision was inelastic by combining the two masses. Again this led to an incorrect result.

What you need to use is your conservation of momentum equation from your relevant equations and one more conservation equation that applies to elastic collisions. Then you will have two equations in two unknowns. Solve for the velocities.

Typically this sort of problem uses conservation of kinetic energy as the second equation. The masses and their velocities must be kept separate: the masses do not stick together. Of course, dealing with the squares of the velocities is annoying algebraically when you try to solve the two equations.

Fortunately there is another equivalent conservation rule which applies to elastic collisions which makes the algebra much easier. The rule is: The closing speed of the two bodies before the collision is equal to their separation velocity after the collision. To make this more succinct and using your variables: v1 - v2 = v2' - v1' .

So, try again using the purely elastic collision paradigm.