Two body collision with a spring

[Agent M27]

Homework Statement

A block of mass m₁ = 1.6 kg is initially moving to the right with a speed of 4 ms^-1 on a frictionless horizontal track and collides with a spring attached to a second block of mass m₂ = 2.1 kg initially moving to the left at a speed of 2.5 ms^-1. The spring constant is 600 NM^-1.

a. Find the velocities of the two blocks after the collision.

Homework Equations

(1) m₁v_1i+m₂v_2i = m₁v_1f+m₂v_2f

(2) v_1i - v_2i = -(v_1f - v_2f)

The Attempt at a Solution

So this is a problem in the book that is worked out for me, but I cannot seem to figure out the reasoning behind the way they combined and manipulated the equations. They first begin by subbing in the known values into EQ (1) which I agree with. They then state to use EQ (2) due to the collision being ellastic, which I don't totally understand this relationship. They sub known values into EQ (2) giving the following:

6.5 ms^-1 = -v_1f = v_2f

The book then states that I must multiply the EQ I gave right above by 1.6 kg which gives me the following which I will label as EQ (3): This is what I don't understand, why would I multiply EQ (2) by 1.6kg?

10.4kgms^-1 = (-1.6kg)v_1f + (1.6kg)v_2f

The book goes on to state that I now must add EQ (1) and EQ (3) which apparently allows me to solve for v_2f, but I don't see where the thought process determines that I add these two EQ. Thanks in advance for the assistance, I am hoping this is a relatively simple algebraic procedure that I am missing.

Joe

 

[Doc Al]

They then state to use EQ (2) due to the collision being ellastic, which I don't totally understand this relationship.

The point here is that EQ (2) is only valid for elastic collisions. One derives that equation by combining EQ (1) with conservation of energy. (The derivation should be in your textbook.) Working with EQ (2) is much easier than working the conservation of energy equation directly.

The book then states that I must multiply the EQ I gave right above by 1.6 kg which gives me the following which I will label as EQ (3): This is what I don't understand, why would I multiply EQ (2) by 1.6kg?

10.4kgms^-1 = (-1.6kg)v_1f + (1.6kg)v_2f

The book goes on to state that I now must add EQ (1) and EQ (3) which apparently allows me to solve for v_2f, but I don't see where the thought process determines that I add these two EQ. Thanks in advance for the assistance, I am hoping this is a relatively simple algebraic procedure that I am missing.

It's just a trick. One equation has a +1.6v₁ term and the other has a -1.6v₁ term. Add them together and the v₁ terms cancel leaving only v₂.