Question

A proton of mass m undergoes a head on collision with a stationary atom of mass 13m. If the initial speed of the proton is 485 m/s, find the speed of the proton after the collision.

Answer 1

485 m = m u + 13 m v
now you have to know if the collision is elastic or if they proton sticks to the atom
If they stick
then u = v
if elastic
then (1/2) m(485^2) = (1/2) m u^2 + (1/2)(13 m) v^2

Answer 2

Given:
M1 = m, V1 = 485 m/s.
M2 = 13m, V2 = 0.
V3 = Velocity of proton(M1) after collision.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
m*485 + 13m*0 = m*V3 + 13m*V4,
485m + 0 = m*V3 + 13m*V4,
Divide both sides by m:
485 = V3 + 13V4,

V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (485(m-13m) + 26m*0)/14m = -5820m /14m = -416 m/s.