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Mechanics

Conservation of Linear Momentum

Consider a 1-dimensional collision between two bodies in which no external forces act.

 

FAB is the force exerted by body A on body B.

FBA is the force exerted by body B on body A.

Newton’s third law of motion states that FBA = - FAB

If the bodies are in contact for a short time, DELTA01t, then

change in momentum of A is DELTA02pA = FBA × DELTA02t = - FAB × DELTA02t

change in momentum of B is DELTA02pB = FAB × DELTA02t

So the total change in momentum (DELTA01pA + DELTA01pB) is zero.

Conclusion

The principle of conservation of momentum is, in effect, an expression of Newton’s third law of motion.

Elastic Collisions: An Example

Two bodies A and B, have an elastic collision during which no external forces act (this can easily be arranged by having magnets with similar poles facing each other, as shown below). Calculate the magnitudes and senses of the velocities of the bodies after the collision.

 

Mass of A = mA = 2 kg

Mass of B = mB = 3 kg

The principle of conservation of momentum can be used

Total momentum 
before collision

=

Total momentum 
after collision
mAuA + mBuB = mAvA + mBvB

 

which gives

2vA + 3vB = 14 

(equation 1)

As the collision is elastic we can use the fact that the total K.E. is the same before and after the collision.

This gives us:

½mAuA² + ½mBuB²

½mAvA² + ½mBvB²

and these two equations can be used to calculate vA and vB.

However, the calculation can be simplified if we remember that for an elastic collision, the relative velocity of approach is equal to the relative velocity of separation.

Velocity of A relative to B before the collision

is equal but opposite to

Velocity of A relative to B after the collision

velocity of A relative to B before collision is

uA – uB = +2 ms-1

velocity of A relative to B after collision is -2 ms-1

therefore 

vA – vB = -2 ms-1

using this and equation 1 allows us to calculate vA and vB easily.

Answers:

vA = +1·6 ms-1

vB = +3·6 ms-1

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