Double Collision of Mass–Spring System

A ball of mass M moving with velocity V₀ on a frictionless plane strikes the first of two identical balls, each of mass m = 2kg connected by a massless spring with spring constant k = 1kg/s² (see Figure 1.1). Consider the collision to be central and elastic and essentially instantaneous.

Figure 1.1

a) Find the minimum value of the mass M for the incident ball to strike the system of two balls again.

b) How much time will elapse between the two collisions?

SOLUTION

a) Let us call the ball of mass M ball 1, the first ball struck 2, and the third ball 3 (see Figure 1.2a). After the first collision, ball 1 will move with constant velocity V₁ and balls 2 and 3 will oscillate. For another collision to take place, the coordinates of balls 1 and 2 must coincide at some later moment. First find V₁ after the initial collision, considered to be instantaneous. Then, this problem is no different from the collision of just two balls of masses M and m. If the velocity of the first ball before the collision is V₀ we can find V₁ and V₂ from energy and momentum

Figure 1.2a

conservation:

Again,

(1)

(2)

After the collision the first ball will move with constant velocity V₁ and so its position coordinate x₁ = V₁t = V₀t(1- γ)/(1+ γ). The center of mass of balls 2 and 3 will also move with constant velocity V_c = V₂/2 (since m₂ = m₃ = m) Therefore from (2)

Now, in the center of mass frame of balls 2 and 3, the two balls are moving toward one another each with speed V₂/2 and they will start to oscillate relative to the center of mass with the following time dependence:

where and k' is the spring constant of half of the spring, k' ≡ 2k. From energy conservation, the initial energy of mass 2 in the center of mass frame goes into the energy of spring deformation with an amplitude corresponding to the velocity change from V₂/2 to zero:

In the lab frame

Figure 1.2b

For the second collision to occur, we need x₁ = x₂ or

(3)

So we have

(4)

The easiest way is to solve (4) graphically (see Figure 1.2b). For the solution to exist, we have the condition γ ≤ γ_max, where

at φ₀ ≈ 3π/2. The minimum value of the mass M = m/γ_max ≈ 10 kg

b) The time between collisions is

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