Two particles, of mass m and mass nm, undergo a head-on, fully elastic collision

Question

Two particles, of mass m and mass nm, undergo a head-on, fully elastic collision. (Here n is just a unitless number.) Before colliding the particles approach one another with equal speed v. Snapshots of the particles before, during, and after the collision are shown above. After the collision the first particle moves away with speed 2.95v in the exact opposite direction, and the velocity of the second particle is unknown. What is the value of n?

Before Collision (m)-><-(nm) Collision ()() After Collision <-(m v=2.95) <-(nm)-> Vfinal=?

Explanation / Answer

Let's say particle nm has a velocity v2 after collision, Since the particles are moving in the opposite direction,

From Conservation of kinetic energy and linear momentum , we have-

nmv-mv=m(2.95v)-nmv2------------------------------(1)

and from conservation of energy

1/2mv^2 +1/2 nm *v^2 = 1/2 m* (2.95v)^2 +1/2 *nm* (v2)^2-----(2)

Solving equation 1 and 2,

(3.95-n)v=nv2---(3)

And (n-7.7)v^2=n(v2)^2---(4)

we get,

Dividing 4 by 3

we get, v2= (n-7.7)v/(3.95-n)----(5)

and from equation (3) itself-

v2=(3.95-n)v/n---(6)

So solving (5) and (6), two equations and two variables,n and v2

we get, n=79

and v2= -0.95v( body nm will move in same drection as the body m)

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