Question

Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on. Find the subsequent speed of mass, m and 4m.

Answer 1

First mass (M1)= m, speed (V1) =2v Second mass(M2)= 4m, speed(V2)= v PRINCIPLE OF CONSERVATION OF MOMENTUM SAYS M1*V1+M2*V2= M1*V1f +M2*V2f where f = final speed finally speed is assumed same, therefore V1f=V2f =Vf m*2v +4m*v = (m+4m)Vf 6vm = Vf(5m) dividing both sides by 5m 6/5v=Vf 1.2v=Vf = final speed

Answer 2

Refer to the figure shown below.

Let
v₁ and v₂ = velocities of masses m₁, m₂ after the collision, respectively.

Note:
In an elastic collision, both momentum and kinetic energy are conserved.

For conservation of momentum,
m(2v) + 4m(v) = m(v₁) + 4m(v₂)
6v = v₁ + 4v₂ (1)

For conservation of kinetic energy,
(1/2)m(2v)² + (1/2)4m(v²) = (1/2)m(v₁²) + (1/2)4m(v₂²)
8v² = v₁² + 4v₂² (2)

Substitute (1) into (2).
8v² = (6v - 4v₂)² + 4v₂²
= 36v² - 48 v₂v + 16v₂² + 4v₂²
20v₂² - 48v v₂ + 28v² = 0
v₂² - 2.4v v₂ + 1.4v² = 0
v₂ = 0.5[2.4v +/- v√(0.16)]
v₂ = 1.6v, or 0.8v
Correspondingly,
v₁ = -0.4v, or 2.8v

If KE were not conserved, the mass m should rebound with negative velocity.
Furthermore, it is unlikely that mass 4m would attain a velocity of 2.8v.
Therefore the acceptable solution is
v₁ = -0.4v and v₂ = 1.6v

Answer:
After the collision,
The mass m has velocity of -0.4v,
The mass 4m has velocity of 1.6v.