MB3, MB4 - Impulse

MB3, MB4 - Impulse

Impulse measures the accumulated effect of a force and equals the change in momentum. You will retain its vector direction, connect constant or average force with contact time, and integrate a variable force in one dimension using signed areas.

Impulse as vector momentum change

A 0.2 kg0.2\,\mathrm{kg} ball travels right at 10 m s−110\,\mathrm{m\,s}^{-1}, then rebounds left at 8 m s−18\,\mathrm{m\,s}^{-1}. Is its momentum change small because its speed changes by only 2? Its velocity changes from +10+10 to −8-8, so the change is −18 m s−1-18\,\mathrm{m\,s}^{-1}, not −2-2.

The impulse of the resultant force over an interval equals the body's momentum afterwards minus its momentum beforehand. For the opening ball it is 0.2(−8−10)=−3.6 kg m s−10.2(-8-10)=-3.6\,\mathrm{kg\,m\,s}^{-1}, directed left. A vector change includes any change in direction, even when speed stays constant.

Impulse–momentum relation

For a particle of constant mass mm,

I=mv−mu=m(v−u)\mathbf I=m\mathbf v-m\mathbf u=m(\mathbf v-\mathbf u)

This is the impulse of the resultant force between the two velocity measurements. It is the impulse of a particular contact force only if other impulses are negligible or have been accounted for. The unit is N s\mathrm{N\,s}, equivalent to kg m s−1\mathrm{kg\,m\,s}^{-1}.

Not in the formula booklet: learn it.

In a plane, subtract components before finding a magnitude. The magnitude ∣I∣|\mathbf I| is nonnegative, but an impulse component may be negative.

Worked example 1: changing one velocity component

Question. A 0.2 kg0.2\,\mathrm{kg} particle changes velocity from 3i+4j3\mathbf i+4\mathbf j to 3i−2j3\mathbf i-2\mathbf j, in m s−1\mathrm{m\,s}^{-1}. Find the resultant impulse vector and its magnitude.

Plan. Find final minus initial velocity component by component, then multiply by mass.

I=0.2[(3i−2j)−(3i+4j)]=−1.2j N s∣I∣=1.2 N s\begin{aligned} \mathbf I&=0.2\bigl[(3\mathbf i-2\mathbf j)-(3\mathbf i+4\mathbf j)\bigr]\\ &=-1.2\mathbf j\,\mathrm{N\,s}\\ |\mathbf I|&=1.2\,\mathrm{N\,s} \end{aligned}

Answer. Impulse is −1.2j N s-1.2\mathbf j\,\mathrm{N\,s}, magnitude 1.2 N s1.2\,\mathrm{N\,s}.

Check. Initial momentum is 0.6i+0.8j0.6\mathbf i+0.8\mathbf j; adding the impulse gives 0.6i−0.4j=0.2(3i−2j)0.6\mathbf i-0.4\mathbf j=0.2(3\mathbf i-2\mathbf j), the required final momentum.

Change only the final horizontal component from +3+3 to −3-3. The impulse now has two nonzero components.

Worked example 2: a two-component impulse

Question. The same particle changes from 3i+4j3\mathbf i+4\mathbf j to −3i−2j-3\mathbf i-2\mathbf j, in m s−1\mathrm{m\,s}^{-1}. Find the resultant impulse vector and its magnitude.

Plan. Subtract the full initial vector, then take the length of the resulting impulse vector.

I=0.2[(−3−3)i+(−2−4)j]=(−1.2i−1.2j) N s∣I∣=1.22+1.22=1.22 N s\begin{aligned} \mathbf I&=0.2\bigl[(-3-3)\mathbf i+(-2-4)\mathbf j\bigr]\\ &=(-1.2\mathbf i-1.2\mathbf j)\,\mathrm{N\,s}\\ |\mathbf I|&=\sqrt{1.2^2+1.2^2}=1.2\sqrt2\,\mathrm{N\,s} \end{aligned}

Answer. I=(−1.2i−1.2j) N s\mathbf I=(-1.2\mathbf i-1.2\mathbf j)\,\mathrm{N\,s}, magnitude 1.22 N s1.2\sqrt2\,\mathrm{N\,s}.

Check. Adding this impulse to the initial momentum gives −0.6i−0.4j-0.6\mathbf i-0.4\mathbf j, exactly the stated final velocity times 0.20.2. Its equal negative components point diagonally down and left.

During a two-body collision, the contact impulse on BB is the negative of that on AA, by Newton's third law. Say which body experiences your impulse. That sign change matters when a later question asks for the force exerted on the other body.

Watch out

Wrong: ∣I∣=m(∣v∣−∣u∣)|\mathbf I|=m(|\mathbf v|-|\mathbf u|). A change of speed omits direction. Use m(v−u)m(\mathbf v-\mathbf u), then take the vector magnitude if requested.

Apply the vector subtraction to new data.

Constant and average force

For constant resultant force, Newton's second law gives F=ma\mathbf F=m\mathbf a. Over a time Δt\Delta t, constant acceleration gives v−u=aΔt\mathbf v-\mathbf u=\mathbf a\Delta t. Combining them yields FΔt=m(v−u)\mathbf F\Delta t=m(\mathbf v-\mathbf u).

Force and duration

For a constant resultant force over duration Δt>0\Delta t>0,

I=FΔt\mathbf I=\mathbf F\Delta t

For a varying force, its average over the interval is defined by

Fav=IΔt\mathbf F_{\mathrm{av}}=\frac{\mathbf I}{\Delta t}

Thus I=FavΔt\mathbf I=\mathbf F_{\mathrm{av}}\Delta t, even when the instantaneous force is not constant. This does not mean the force equals its average at every instant.

Not in the formula booklet: learn it.

For a fixed momentum change, shortening the contact time increases the magnitude of average force. This explains why extending a stopping time can reduce average force without changing the impulse needed to stop the object.

Worked example 3: force during a rebound

Question. A 0.5 kg0.5\,\mathrm{kg} ball changes velocity from +4+4 to −2 m s−1-2\,\mathrm{m\,s}^{-1} in 0.10 s0.10\,\mathrm{s}, right positive. Other impulses are negligible. Find the average contact force on the ball.

Plan. Find the signed impulse, then divide by contact duration.

I=0.5(−2−4)=−3 N sFav=−30.10=−30 N\begin{aligned} I&=0.5(-2-4)=-3\,\mathrm{N\,s}\\ F_{\mathrm{av}}&=\frac{-3}{0.10}=-30\,\mathrm{N} \end{aligned}

Answer. Average force is 30 N30\,\mathrm{N} leftwards. If the contact force is assumed constant, it is −30 N-30\,\mathrm{N} throughout the interval.

Check. The average-force rectangle has signed area (−30)(0.10)=−3 N s(-30)(0.10)=-3\,\mathrm{N\,s}, matching the momentum change.

Keep the mass and velocities but shorten only the contact duration by a factor of five.

Worked example 4: the same impulse in less time

Question. The same change from +4+4 to −2 m s−1-2\,\mathrm{m\,s}^{-1} now takes 0.020 s0.020\,\mathrm{s}. Find the average contact force on the 0.5 kg0.5\,\mathrm{kg} ball.

Plan. Keep the impulse −3 N s-3\,\mathrm{N\,s} and divide by the new time.

Fav=−30.020=−150 NF_{\mathrm{av}}=\frac{-3}{0.020}=-150\,\mathrm{N}

Answer. Average force is 150 N150\,\mathrm{N} leftwards.

Check. The time is one fifth as long and force magnitude is five times as large; their product remains −3 N s-3\,\mathrm{N\,s}. Under a constant-force model, a=−150/0.5=−300 m s−2a=-150/0.5=-300\,\mathrm{m\,s}^{-2}, and 4+(−300)(0.020)=−24+(-300)(0.020)=-2.

The same equations apply component by component. For example, impulse (2i−j) N s(2\mathbf i-\mathbf j)\,\mathrm{N\,s} over 0.25 s0.25\,\mathrm{s} gives average force (8i−4j) N(8\mathbf i-4\mathbf j)\,\mathrm{N}. The partner body receives the opposite contact force.

Watch out

Wrong: F=IΔtF=I\Delta t. Impulse is force times time, so force is impulse divided by time. Check units: N s/s=N\mathrm{N\,s}/\mathrm{s}=\mathrm{N}, not N s2\mathrm{N\,s}^2.

Find a fresh vector average force.

Integrating a variable force in one dimension

A changing force cannot usually be replaced by its value at the end of the interval. Divide time into short intervals: over each one, force is approximately constant, giving a small impulse FΔtF\Delta t. Adding these and taking the limiting sum gives an integral.

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