M1 4.3 - Momentum and impulse

M1 4.3 - Momentum and impulse

Track direction as well as speed when a force acts briefly or two particles collide. You will calculate momentum and impulse, conserve total momentum when external impulse is negligible, and interpret a negative velocity after impact.

Momentum with signs

Can a collision leave one particle moving backwards even though the pair's total momentum is forwards? Momentum lets us answer this without trying to follow the rapidly changing contact forces.

Choose one positive direction along the line of motion. Momentum is mass multiplied by velocity, so it has direction. Mass is positive; the velocity supplies the sign.

Momentum

p=mv.p=mv.

Momentum has units kg m s−1\mathrm{kg\,m\,s^{-1}}. For several particles on one line, add their signed momenta. Not in the formula booklet: learn it.

First calculate the momentum of one ball, keeping the chosen direction visible.

Worked example 1: moving right

A 0.6 kg0.6\,\mathrm{kg} ball moves right at 5 m s−15\,\mathrm{m\,s^{-1}}. Find its momentum, taking right as positive.

Plan. Use positive velocity for motion in the chosen positive direction.

p=0.6(5)=3 kg m s−1.p=0.6(5)=3\,\mathrm{kg\,m\,s^{-1}}.

Answer. 3 kg m s−13\,\mathrm{kg\,m\,s^{-1}} rightwards.

Check. Dividing the momentum by 0.6 kg0.6\,\mathrm{kg} recovers velocity +5 m s−1+5\,\mathrm{m\,s^{-1}}.

Change only the ball's direction. Its speed is unchanged.

Worked example 2: moving left

The same ball moves left at 5 m s−15\,\mathrm{m\,s^{-1}}. Find its momentum, still taking right as positive.

Plan. The velocity is now −5 m s−1-5\,\mathrm{m\,s^{-1}}.

p=0.6(−5)=−3 kg m s−1.p=0.6(-5)=-3\,\mathrm{kg\,m\,s^{-1}}.

Answer. Signed momentum −3 kg m s−1-3\,\mathrm{kg\,m\,s^{-1}}, or momentum of magnitude 3 kg m s−13\,\mathrm{kg\,m\,s^{-1}} leftwards.

Check. Adding this to the momentum of an identical ball moving right gives zero total momentum, although both balls are moving.

Zero total momentum does not mean zero total speed: opposite signed momenta can cancel.

Impulse as change in momentum

A brief force can produce a large change in momentum. Its impulse is the change in momentum it produces, with initial and final velocities measured using the same positive direction.

Impulse–momentum principle

I=mv−mu=m(v−u).I=mv-mu=m(v-u).

For a constant force, I=FΔtI=F\Delta t; for a varying force, I=FavΔtI=F_{\mathrm{av}}\Delta t. Impulse has units N s\mathrm{N\,s}, equivalent to kg m s−1\mathrm{kg\,m\,s^{-1}}. Not in the formula booklet: learn it.

The force in this relation is the resultant force. In a very brief impact we commonly neglect other impulses, such as weight's, compared with the contact impulse; state or use the model given in the problem.

Worked example 3: a rebound

A 0.2 kg0.2\,\mathrm{kg} ball travelling right at 8 m s−18\,\mathrm{m\,s^{-1}} hits a wall and rebounds left at 6 m s−16\,\mathrm{m\,s^{-1}}. Contact lasts 0.04 s0.04\,\mathrm s; other impulses are negligible. Find the impulse and average contact force on the ball.

Plan. Take right positive, so the final velocity is negative.

I=0.2(−6−8)=−2.8 N s,Fav=−2.80.04=−70 N.I=0.2(-6-8)=-2.8\,\mathrm{N\,s}, \qquad F_{\mathrm{av}}=\frac{-2.8}{0.04}=-70\,\mathrm N.

Answer. Impulse 2.8 N s2.8\,\mathrm{N\,s} leftwards and average force 70 N70\,\mathrm N leftwards.

Check. Initial momentum is +1.6+1.6, final momentum is −1.2-1.2, and −1.2−(+1.6)=−2.8-1.2-(+1.6)=-2.8.

Change only the final direction: suppose instead that a brief braking force slows the ball to 6 m s−16\,\mathrm{m\,s^{-1}} rightwards in the same time.

Worked example 4: slowing without reversal

Find the impulse and average resultant force for this changed final velocity.

Plan. Both velocities are positive; subtract them in final-minus-initial order.

I=0.2(6−8)=−0.4 N s,Fav=−0.4/0.04=−10 N.I=0.2(6-8)=-0.4\,\mathrm{N\,s}, \qquad F_{\mathrm{av}}=-0.4/0.04=-10\,\mathrm N.

Answer. Impulse 0.4 N s0.4\,\mathrm{N\,s} leftwards and average force 10 N10\,\mathrm N leftwards.

Check. The force is opposite the motion, as required to slow the ball. A reversal requires the larger momentum change seen in Example 3.

Subtracting the two speeds in a rebound loses the direction change. Conversely, an impulse's magnitude is always nonnegative.

Conservation for direct collisions

During a collision the particles exert equal and opposite forces on each other, so their mutual impulses cancel when we consider the pair. If the external impulse along the line is negligible, total momentum is unchanged.

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