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Chapter 7 Impulse and Momentum.

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Presentation on theme: "Chapter 7 Impulse and Momentum."— Presentation transcript:

1 Chapter 7 Impulse and Momentum

2 1) Definition p = mv

3 2) Impulse For a constant force, From the 2nd law,
If the force is not constant, use the average force

4 7) Conservation of momentum in 2d
If initial conditions are known, this gives 2 equations, with 4 unknowns, so more information is needed. v1’ v2’

5 7) Conservation of momentum in 2d
a) Inelastic collision: p1’ = p2’ = p’ reduces unknowns to px’ and py’ Example: Find v’ if m1 = 1450 kg, m2 = 1750 kg, v1 = 11.5 m/s, v2 = 15.5 m/ss • conservation of x momentum m1 m2 v1 v2 v’ x y • conservation of y momentum

6 7) Conservation of momentum in 2d
b) Elastic collision: Energy conservation adds 3rd equation: The last condition is determined by the shape & location of impact: e.g. F For a billiard ball collision, the angle of the object ball is determined by the line through the centres at the point of contact.

7 7) Conservation of momentum in 2d
b) Elastic collision: Energy conservation adds 3rd equation: Conservation of momentum v1 v2’ v1’ F 1 Example: Cue ball angle: • identical masses • elastic collision • m2 initially at rest 2? Conservation of energy v2’ v1’ v1 Therefore, by Pythagoras

8 8) Centre of Mass Acceleration and force:
The centre-of-mass of a system of particles (or 3d object) reacts to the sum of the forces like a point particle with a mass equal to the total mass of the particles. If the total force is zero, the centre-of-mass does not accelerate. (If there are no external forces, the internal forces sum to zero by the 3rd law.)

9 8) Centre of Mass b) Momentum and velocity Since a = v/t,
But, since vCM = 0 if v1 = 0 and v2 = 0, this becomess The numerator represents the total momentum, which is conserved in the absence of external forces, so again, the centre-of-mass velocity is constant. Note that the CM momentum is simply equal to the total momentum:

10 8) Centre of Mass c) Position of the centre of mass
Since v = x/t, the above gives If 2 particles coincide, they also coincide with the CM, so In one dimension, For m1 = 5.0 kg, m2 = 12 kg, x1 = 2.0 m, and x2 = 6.0 m, xCM = 4.8 m


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