How to work this physics problem?

Question

A proton (mass 1 u) moving at 7.40 x 10^6 m/s collides elastically and head-on with a second particle moving in the opposite direction at 2.60 x 10^6 m/s. After the collision, the proton is moving opposite to its initial direction at 6.20 x 10^6 m/s. Find the mass and final velocity of the second particle. [Take the proton's initial velocity to be in the positive direction.]
_______________m/s

I managed to get the mass of the second particle which is 2.125 u, but I can't seem to get the velocity, help?

Additional Details

Lab? What are you talking about?

Answer 1

Best Answer - Chosen by Asker

Use the law of conservation of momentum.

M1V1 + M2V2 = M1V3 + M2V4

where

M1 = 1 µ
V1 = 7.4 x 10^6 m/sec.
M2 = mass of the 2nd particle (unknown)
V2 = - 2.6 x 10^6 m/sec.
V3 = - 6.2 x 10^6 m/sec.
V4 = final velocity of the second particle (unknown)

Substituting values,

(1)(7.4 x 10^6) + M2(-2.6 x 10^6) = (1)(-6.2 x 10^6) + M2V4

(7.4 + 6.2) x 10^6 = M2V4 + 2.6 x 10^6(M2)

13.6 x 10^6 = M2V4 + 2.6 x 10^6(M2) --- call this Equation 1

Since the collision is ELASTIC, then the kinetic energy in the system is conserved. Thus being said,

(1/2)(M1)(V1)^2 + (1/2)(M2)(V2)^2 = (1/2)(M1)(V3)^2 + (1/2)(M2)(V4)^2

Simplifying,

(M1)(V1)^2 + (M2)(V2)^2 = (M1)(V3)^2 + (M2)(V4)^2 --- Equation 2

I will leave the rest of the solution for you to do. At this point, you have two equations with two unknowns (M2 and V4). I have done the Physics part and all you have to do is solve these two equations simultaneously.

Hope this helps.

Asker's Rating:

Asker's Comment:

I'll try to figure it out

Answer 2

use the lab

Answer 3

Search for 'Conservation of momentum'