Elastic collision question help?

Question

The question is that two objects make a head on collision and one is moving to the right which is assumed to be positive while the second is moving to the left. The first object is 4.9g with 12cm/s and it makes a collision with a 15g object moving at 15cm/s. I have to find the velocity of the first object after the collision. What I tried was px: m1u1-m2u2=m1v1+m2v2 and u2-u1=-v2+v1 what I got was 4.9*12+15*15=283.8 or -166.2 if it's moving to the left. Then I tried 4.9v1 + 15(v1-3)=283.8 or -166.2. Then I isolated to get v1 but I got the answer wrong for both cases. I just can't get the final speed and I don't know what else to try. I don't think I can use m1u1-m2u2=m1v1+m2v2 because I have two unknowns v1 and v2.

Answer 1

Your first equation, conservation of momentum, has a minus sign where it should have a plus sign. The other equation you need is conservation of kinetic energy: the sum of the 2 original energies equals the sum of the 2 final energies. Then solve the system of equations.

Answer 2

There is a useful trick to solving the equations of Conservation of Energy and of Momentum simultaneously.

Conservation of momentum before and after a collision requires that

m₁v₀₁ + m₂v₀₂  =  m₁v₁ + m₂v₂     [eq. 1]

which can be rearranged as

m₁v₀₁ - m₁v₁ = m₂v₂ - m₂v₀₂,   or

m₁ (v₀₁ - v₁) = m₂ (v₂ - v₀₂). [eq. 2]

A collision takes place so quickly that there are no significant changes in potential energy, and it is often assumed that kinetic energy is conserved as well:

½ m₁v₀₁² + ½ m₂v₀₂²  =  ½ m₁v₁² + ½ m₂v₂²     [eq. 3]

Since ½ occurs in every term, these may be cancelled. Rearranging this equation leads to:

m₁v₀₁² - m₁v₁²  =  m₂v₂² - m₂v₀₂², or

m₁ (v₀₁² - v₁²)  =  m₂ (v₂² - v₀₂²).     [eq. 4]

The reason for these rearrangements is the simple result that comes when eq. 2 is divided into eq. 4:

m₁(v₀₁² - v₁²)/m₁(v₀₁ - v₁) = m₂(v₂² - v₀₂²)/m₂(v₂ - v₀₂)   [eq.5]

since (v₀₁² - v₁²) = (v₀₁ - v₁) (v₀₁ + v₁),

and (v₂² - v₀₂²) = (v₂ - v₀₂) (v₂ + v₀₂), eq. 5 becomes

v₀₁ + v₁ = v₂ + v₀₂ , or

v₁ - v₂  =  v₀₂ - v₀₁.     [eq. 6]

It is generally easier to use eq. 2 and eq. 6, to solve problems such as yours than eq. 1 and eq. 3, which are more cumbersome.

NOTE: In the real world there is a further complication, although it did not arise in your problem. Not all collisions are 'elastic,' meaning that kinetic energy is never exactly conserved, because heat losses are always present. To compensate for this, eq. 6 is further revised to

(v₁ - v₂) / (v₀₂ - v₀₁) = e,

where e, the coefficient of restitution, is equal to 1 for an elastic collision, but e is less than 1 for real collisions. There is the further complication that there are no established values for e, which varies with the materials, their dimensions, and their relative speeds.