Mi1–Mi3 - Impulse and momentum as vectors

Mi1–Mi3 - Impulse and momentum as vectors

Calculate momentum and impulse with their directions intact. You will use the total external impulse to find a velocity change, handle a rebound and a mean contact force, and use relative velocity to describe approach and separation along a line.

Momentum and force impulse

A 0.15 kg0.15\,\mathrm{kg} ball strikes a wall at 8 m s−18\,\mathrm{m\,s^{-1}} and returns at the same speed. Has its momentum changed?

Its speed has not changed, but its direction has reversed. Momentum records that distinction. By the third section you will calculate the wall's impulse without needing to know how the force varied during contact.

You already know

Velocity is a vector; speed is its magnitude. The perpendicular unit vectors i\mathbf{i}, j\mathbf{j}, k\mathbf{k} describe three coordinate directions. The magnitude of ai+bj+cka\mathbf{i}+b\mathbf{j}+c\mathbf{k} is a2+b2+c2\sqrt{a^2+b^2+c^2}. Divide a non-zero direction vector by its magnitude to make a unit vector.

More mass at the same velocity means more momentum. A force acting for longer supplies more impulse. Both quantities point in a direction, but they describe different things: a body's motion and a force's accumulated effect.

Momentum and impulse

The linear momentum of a body is its mass multiplied by its velocity. The impulse of a force measures its effect over a time interval. Both are vectors: momentum points with the velocity, and the impulse of a constant force points with that force.

Their units look different but are equivalent: 1 N s=1 kg m s−11\,\mathrm{N\,s}=1\,\mathrm{kg\,m\,s^{-1}}.

Momentum and constant-force impulse

For mass mm, velocity v\mathbf{v}, and a constant force F\mathbf{F} acting for time Δt\Delta t,

p=mv,J=FΔt\mathbf{p}=m\mathbf{v},\qquad \mathbf{J}=\mathbf{F}\Delta t

Use kilograms, metres per second, newtons and seconds for SI values.

Not in the formula booklet: learn it.

Multiplying a vector by a positive mass or duration changes its length, not its direction.

Worked example 1: two different vector quantities

Question. A particle of mass 0.5 kg0.5\,\mathrm{kg} initially has velocity (6i−8j) m s−1(6\mathbf{i}-8\mathbf{j})\,\mathrm{m\,s^{-1}}. A constant force (4i+3j) N(4\mathbf{i}+3\mathbf{j})\,\mathrm{N} acts for 2 s2\,\mathrm{s}. Find its initial momentum and the impulse of this force.

Plan. Multiply the initial velocity by the mass, and the force by its duration.

p=0.5(6i−8j)=3i−4jmass times velocityJ=2(4i+3j)=8i+6jduration times force\begin{aligned} \mathbf{p}&=0.5(6\mathbf{i}-8\mathbf{j}) =3\mathbf{i}-4\mathbf{j} && \text{mass times velocity} \\ \mathbf{J}&=2(4\mathbf{i}+3\mathbf{j}) =8\mathbf{i}+6\mathbf{j} && \text{duration times force} \end{aligned}

Why this step? The mass belongs in the momentum calculation; the force duration belongs in the impulse calculation.

Answer. Initial momentum (3i−4j) kg m s−1(3\mathbf{i}-4\mathbf{j})\,\mathrm{kg\,m\,s^{-1}}; impulse (8i+6j) N s(8\mathbf{i}+6\mathbf{j})\,\mathrm{N\,s}.

Check. Their magnitudes are 5 kg m s−15\,\mathrm{kg\,m\,s^{-1}} and 10 N s10\,\mathrm{N\,s}, agreeing with 0.5(10)0.5(10) and 2(5)2(5).

Change only how the same force is supplied: its magnitude and a direction vector are given separately. The direction vector need not have length one, so normalise it first.

Worked example 2: magnitude and a non-unit direction

Question. The same particle initially has velocity (6i−8j) m s−1(6\mathbf{i}-8\mathbf{j})\,\mathrm{m\,s^{-1}}. A force of magnitude 5 N5\,\mathrm{N} acts in direction 4i+3j4\mathbf{i}+3\mathbf{j} for 2 s2\,\mathrm{s}. Find its initial momentum and the force impulse.

Plan. Convert the direction vector to a unit vector before multiplying by the force magnitude.

∣d∣=42+32=5direction-vector lengthF=5(4i+3j5)=4i+3jmagnitude times unit directionp=0.5(6i−8j)=3i−4jinitial momentumJ=2(4i+3j)=8i+6jforce times duration\begin{aligned} |\mathbf{d}|&=\sqrt{4^2+3^2}=5 && \text{direction-vector length} \\ \mathbf{F}&=5\left(\frac{4\mathbf{i}+3\mathbf{j}}5\right) =4\mathbf{i}+3\mathbf{j} && \text{magnitude times unit direction} \\ \mathbf{p}&=0.5(6\mathbf{i}-8\mathbf{j}) =3\mathbf{i}-4\mathbf{j} && \text{initial momentum} \\ \mathbf{J}&=2(4\mathbf{i}+3\mathbf{j}) =8\mathbf{i}+6\mathbf{j} && \text{force times duration} \end{aligned}

Why this step? Multiplying 55 by the raw direction vector would create a force of magnitude 25 N25\,\mathrm{N}, five times too large.

Answer. (3i−4j) kg m s−1(3\mathbf{i}-4\mathbf{j})\,\mathrm{kg\,m\,s^{-1}} and (8i+6j) N s(8\mathbf{i}+6\mathbf{j})\,\mathrm{N\,s}.

Check. The actual force magnitude is 42+32=5 N\sqrt{4^2+3^2}=5\,\mathrm{N}; the impulse magnitude is 5(2)=10 N s5(2)=10\,\mathrm{N\,s}.

A useful surprise: two moving particles can have zero total momentum. Equal masses with opposite velocities have equal and opposite momentum vectors, even though neither particle is at rest.

Watch out

Wrong: replace a vector by its magnitude before multiplying. Right: retain its components. A momentum of magnitude 55 does not tell you whether the body travels left, right or diagonally.

Use the same two calculations with fresh data.

Using total impulse to find velocity

The total external impulse changes a body's momentum, so its final momentum is its initial momentum plus that impulse.

Impulse–momentum equation

For a body of constant mass mm, with initial velocity u\mathbf{u}, final velocity v\mathbf{v}, and total external impulse J\mathbf{J},

J=mv−muorv=u+Jm\mathbf{J}=m\mathbf{v}-m\mathbf{u} \qquad\text{or}\qquad \mathbf{v}=\mathbf{u}+\frac{\mathbf{J}}m

Include the impulse of every external force, or work in a direction where the omitted forces have zero resultant impulse.

Not in the formula booklet: learn it.

Apply the equation component by component, in two or three dimensions. If an impulse is supplied by magnitude and a direction vector, first convert it to components using a unit direction, just as for a force. In a horizontal problem, weight and the normal reaction may cancel vertically, but that cancellation must come from the model rather than from forgetting the forces.

Worked example 3: a three-dimensional impulse

Question. A 2 kg2\,\mathrm{kg} particle has velocity (3i−j) m s−1(3\mathbf{i}-\mathbf{j})\,\mathrm{m\,s^{-1}}. Its total external impulse is (4i+6j−2k) N s(4\mathbf{i}+6\mathbf{j}-2\mathbf{k})\,\mathrm{N\,s}. Find its final velocity and speed.

Plan. Add the impulse to the initial momentum, divide by the mass, then take the three-dimensional magnitude.

2v=2(3−10)+(46−2)momentum plus impulse=(104−2)combine componentsv=(52−1)divide by 2∣v∣=52+22+(−1)2=30speed is magnitude\begin{aligned} 2\mathbf{v}&=2\begin{pmatrix}3\\-1\\0\end{pmatrix} +\begin{pmatrix}4\\6\\-2\end{pmatrix} && \text{momentum plus impulse} \\ &=\begin{pmatrix}10\\4\\-2\end{pmatrix} && \text{combine components} \\ \mathbf{v}&=\begin{pmatrix}5\\2\\-1\end{pmatrix} && \text{divide by }2 \\ |\mathbf{v}|&=\sqrt{5^2+2^2+(-1)^2}=\sqrt{30} && \text{speed is magnitude} \end{aligned}

Why this step? There was no initial k\mathbf{k}-component, but the impulse creates one. The impulse changes momentum, so dividing by mass is still essential.

Answer. Velocity (5i+2j−k) m s−1(5\mathbf{i}+2\mathbf{j}-\mathbf{k})\,\mathrm{m\,s^{-1}}; speed 30 m s−1\sqrt{30}\,\mathrm{m\,s^{-1}}.

Check. 2[(5,2,−1)−(3,−1,0)]=(4,6,−2)2[(5, 2, -1)-(3, -1, 0)]=(4, 6, -2), recovering every impulse component.

Change only the mass to 1 kg1\,\mathrm{kg}. The same impulse produces twice the velocity change.

Worked example 4: changing the mass

Question. A 1 kg1\,\mathrm{kg} particle has the same initial velocity and receives the same total impulse. Find its final velocity and speed.

Plan. Use the new mass in the impulse–momentum equation.

v=(3−10)+(46−2)m=1=(75−2)add components∣v∣=72+52+(−2)2=78magnitude\begin{aligned} \mathbf{v}&=\begin{pmatrix}3\\-1\\0\end{pmatrix} +\begin{pmatrix}4\\6\\-2\end{pmatrix} && m=1 \\ &=\begin{pmatrix}7\\5\\-2\end{pmatrix} && \text{add components} \\ |\mathbf{v}|&=\sqrt{7^2+5^2+(-2)^2}=\sqrt{78} && \text{magnitude} \end{aligned}

Why this step? Each velocity-change component doubles when the mass halves, not just the component along the original motion.

Answer. (7i+5j−2k) m s−1(7\mathbf{i}+5\mathbf{j}-2\mathbf{k})\,\mathrm{m\,s^{-1}}, with speed 78 m s−1\sqrt{78}\,\mathrm{m\,s^{-1}}.

Check. 1[(7,5,−2)−(3,−1,0)]=(4,6,−2)1[(7, 5, -2)-(3, -1, 0)]=(4, 6, -2).

Always distinguish velocity from speed in the final sentence. A correct vector does not yet answer a request for speed.

Watch out

Wrong: v=u+J\mathbf{v}=\mathbf{u}+\mathbf{J} for every mass. Right: v=u+J/m\mathbf{v}=\mathbf{u}+\mathbf{J}/m. Impulse and velocity have different units.

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