Two particles approach each other with equal and opposite speed, v. The mass of one particle is m, and the mass of the other
particle is nm, where n is just a constant factor.
After the collision, the first particle moves in the exact opposite direction with speed 0.750v, and the speed of the second
particle, Um final, is unknown.
What is the value of n?

Momentum is conserved before and after a collision. If the collision is elastic, then kinetic energy is also conserved. We can use these two equations to solve a system of equations for the unknown variable n.

Conservation of momentum:

p₀ = p

mv + (nm) (-v) = m (-0.750 v) + (nm) (vf)

mv − nmv = -0.750 mv + nm vf

v − nv = -0.750 v + n vf

1.750 v = n (v + vf)

n = 1.750 v / (v + vf)

Without knowing the type of collision, we cannot solve for a numerical value of n. However, if we assume the collision is elastic, then the total kinetic energy is conserved, and we can use that to write a second equation and solve the system of equations for n.

KE₀ = KE

½ mv² + ½ (nm) (-v)² = ½ m (-0.750 v)² + ½ (nm) vf²

½ mv² + ½ nmv² = ½ m (0.5625 v²) + ½ nm vf²

v² + nv² = 0.5625 v² + n vf²

0.4375 v² = n vf² − nv²

To solve the system of equations for n, we must first eliminate the other unknown variable, vf. Start by solving for n vf in the first equation.

n vf = 1.750 v − nv

n vf = v (1.750 − n)

Square both sides:

n² vf² = v² (1.750 − n)²

n² vf² = 3.0625 v² − 3.5 nv² + n²v²

Multiply the second equation by n.

0.4375 nv² = n² vf² − n²v²

Substitute and solve for n:

0.4375 nv² = 3.0625 v² − 3.5 nv² + n²v² − n²v²

0.4375 nv² = 3.0625 v² − 3.5 nv²

0.4375 n = 3.0625 − 3.5 n

3.9375 n = 3.0625

n = 0.778