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Conservation of Momentum of Systems

When two objects A and B collide, the collision can be either (1) elastic or (2) inelastic. Momentum is conserved in all collisions when no external forces are acting. However kinetic energy is conserved in elastic collisions only.

Inelastic Collisions

In inelastic collision, there may be deformations of the object colliding and loss of energy through heat. The heat and the energy to deform the objects comes from the kinetic energy of the objects bfore collison. In inelastic collisons, the momentum is conserved but the kinetic energy is not.

Example 1
On a smooth surface, a soft 100-grams ball A at the velocity of 10 meters per second collides with another 700-grams ball B initially at rest. After collision, the two balls stick together and keep moving in the same direction as ball A. What is the velocity of the two balls after collision?

Solution to Example 1
Let p1 be the momentum of the two balls before collision.
Momentum of ball A: pA = mass × velocity = 0.1 × 10 = 1 Kg.m/s
Momentum of ball B: pB = mass × velocity = 0.7 × 0 = 0 Kg.m/s
p1 = pA + pB = 1 Kg.m/s
After collision the two balls make one ball of mass 0.1 Kg + 0.6 Kg = 0.8 Kg. Let v be the vlocity of the balls after collision. p2 the momentum of the two balls after collison is given by
p2 = 0.8 × v
Momenta are conserved, hence p1 = p2 gives
1 = 0.8 v
v = 1.25 m/s

Elastic Collisions

In elastic collision there are no deformations or transfer of energy in the form of heat and therefore kinetic energy and therefore both momentun and kinetic energy are conserved. A good example is the collision of two billiard balls.

Example 1
On a smooth surface, a soft 100-grams ball A at the velocity of 10 meters per second collides elastically with another 200-grams ball B with velocity 5 meters per second. The velocities of the two balls are in the same direction. After collision, balls A and B keep moving in the same direction at velocities v1 and v2 respectively. What are the velocities v1 and v2?

Solution to Example 1
Let p1 be the momentum of the two balls before collision.
Momentum of ball A: pA = mass × velocity = 0.1 × 10 = 1 Kg.m/s
Momentum of ball B: pB = mass × velocity = 0.2 × 5 = 1 Kg.m/s
p1 = pA + pB = 2 Kg.m/s
p2 the momentum of the two balls after collison is given by
p2 = 0.1 × v1 + 0.2 × v2
Momenta are conserved, hence p1 = p2 gives
2 = 0.1 × v1 + 0.2 × v2
The above is equation with two unknowns: v1 and v2
Since the collision is elastic, there is also conservation of kinetic energy ,hence (using the formula for kinetic energy: (1/2) m v2)
Kinetic energy before collision: K1 = (1/2)(0.1)(10)2 + (1/2)(0.2)(5)2
Kinetic energy after collision: K1 = (1/2)(0.1)(v1)2 + (1/2)(0.2)(v2)2
Conservation: (1/2)(0.1)(10)2 + (1/2)(0.2)(5)2 = (1/2)(0.1)(v1)2 + (1/2)(0.2)(v2)2
Simplify and rewrite the above equation as
15 = 0.1 v12 + 0.2 v22
We now have two equations with two unknowns to solve:
2 = 0.1 × v1 + 0.2 × v2 (equation 1)
and
15 = 0.1 v12 + 0.2 v22 (equation 2)
Equation (2) gives
v1 = 20 - 2 v2
Substitute v1 by 20 - 2 v2 in equation (2) to obtain a quadratic equation in one variable
15 = 0.1 (20 - 2 v2)2 + 0.2 v22
Rearrange and write as
3 v22 - 40 v2 + 125 = 0
Solve to obatain two solutions for v2
v2 = 8.3 m/s and v2 = 5 m/s
Find v1 using v1 = 20 - 2 v2
for v2 = 8.3 m/s , v1 = 20 - 2(8.3) = 3.4 m/s
for v2 = 5 m/s , v1 = 20 - 2(5) = 10 m/s
It first looks like we have two solutions to our problem, but examining the velocities, the solution v2 = 5 m/s and v1 = 10 m/s corresponds to the initial velocities meaning no collision happened and therefore the second set of solutions makes sense.
The velocities after collsion are: v1 = 3.4 m/s and v2 = 8.3 m/s.
Notes: After collision, the velocity of ball A has decreased and that of ball B has increased meaning that part of the kinetic energy of A has been transfered to ball B but this happened with the system of the two balls. Overall the kinetic energy and the momenta before and after collison for the two balls are the same (conserved).

More References
1 - Higher Level Physics - IB Diploma - Chris Hamper - Pearson
2 - Physics - Raymond A. Serway and Jerry S. Faughn
Holt, Reinehart and Winston - Harcourt Education Company


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Updated: 26 September 2015

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