Lesson 3 · 40 min

The Principle of Impulse and Momentum

Start with the momentum a particle has, add the impulse of every force that acts on it over an interval, and you have its momentum at the end. Written in components, that one vector statement finds velocities after a given time, times to reach a velocity, and constant forces, all without the acceleration.

Learning objectives

State the principle of linear impulse and momentum, \(m\vvec_1 + \sum\int\Fvec\,dt = m\vvec_2\), and write it in components.
Draw impulse and momentum diagrams for a particle over an interval.
Apply a four-step procedure to find velocities, times and constant forces, including problems where friction changes direction.
Apply the principle to bodies joined by a cord, one body at a time.

The principle

Lesson 1 integrated Newton's second law in time, and Lesson 2 named both sides. Move the initial momentum to the left:

Principle of linear impulse and momentum

\[ \colL{m\vvec_1} + \textstyle\sum\displaystyle\int_{t_1}^{t_2}\Fvec\,dt = \colL{m\vvec_2} \]

The particle's momentum at \(t_1\), plus the impulses of all the forces acting on it from \(t_1\) to \(t_2\), equals its momentum at \(t_2\).

It is a vector equation. In a plane, write it once for each direction:

\[ m(v_x)_1 + \textstyle\sum\displaystyle\int_{t_1}^{t_2} F_x\,dt = m(v_x)_2 \qquad m(v_y)_1 + \textstyle\sum\displaystyle\int_{t_1}^{t_2} F_y\,dt = m(v_y)_2 \]

Two equations, two unknowns. Typically a velocity and a time, or a velocity and a force such as a normal force.
Choose the axes to suit the problem: along and across an incline, or horizontal and vertical. Use the same positive direction for the momenta and the impulses.
Velocities are measured in an inertial frame, which for these problems is the ground.
Hibbeler writes the principle with \(\sum\int\Fvec\,dt\) between the momenta; Beer & Johnston write \(\text{Imp}_{1\to2}\) for the impulse; Meriam & Kraige write \(\mathbf G = m\vvec\) for the momentum. Same principle.

Impulse and momentum diagrams

The principle has a picture. Draw the particle three times: with its initial momentum, with the impulses of all the forces that act during the interval, and with its final momentum. The vectors in the first two drawings add up to the vector in the third.

Figure 3.1 Impulse and momentum diagrams for a crate pushed across a rough floor. Left: the momentum \(\colL{m\vvec_1}\) at \(t_1\). Middle: the impulse of each force over the interval, drawn like a free-body diagram: the weight's impulse ends at the center of gravity, the floor's push and the applied push end on the faces they push, and friction starts on the bottom face, where it acts. Right: the momentum \(\colL{m\vvec_2}\) at \(t_2\). For constant forces, each impulse is the force times \(\Delta t = t_2 - t_1\).

The middle drawing is the free-body diagram with every force multiplied by the time it acts. That is why the procedure starts with a free-body diagram: a force missing from it is an impulse missing from the equation.

Procedure for analysis

Four steps

Interval and axes. Pick \(t_1\) and \(t_2\): when you know the velocity and when you want it. Choose a positive direction for each axis.
Free-body diagram at a general instant during the interval, then the impulse and momentum diagrams. Check whether any force changes during the interval, or changes direction.
Impulses. \(F\,\Delta t\) for a constant force; \(\int F\,dt\), or the area under its graph, for one that varies (Lesson 4). Find unknown normal forces from the direction in which nothing moves.
Solve and check. One equation per direction. Check the signs, especially for a velocity that reverses, and that friction kept the same direction throughout.

Example 3.1 — An angled push on a rough floor

A worker pushes a \(50\ \text{kg}\) crate, initially at rest, with a constant force \(P = 300\ \text{N}\) directed \(30^\circ\) below the horizontal. The coefficients of friction are \(\mu_s = 0.35\) and \(\mu_k = 0.3\). Find the speed of the crate after \(3\ \text{s}\).

Show solution

Vertical, positive up. The crate stays on the floor, so its vertical momentum stays zero:

\[ 0 + N(3) - 490.5(3) - 300\sin 30^\circ\,(3) = 0 \quad\Rightarrow\quad N = 640.5\ \text{N} \]

Pushing down presses the crate into the floor: \(N\) is larger than the weight.

Does it slide? The horizontal push, \(300\cos 30^\circ = 259.8\ \text{N}\), is more than the most static friction can give, \(\mu_s N = 224.2\ \text{N}\). So it slides from the start, with \(\colF{F_f} = \mu_k N = 192.2\ \text{N}\) against the motion.

Horizontal, positive in the direction of the push:

\[ 0 + \colP{259.8(3)} \colF{{} - 192.2(3)} = 50\,v_2 \quad\Rightarrow\quad 779.4 - 576.5 = 50\,v_2 \quad\Rightarrow\quad v_2 = 4.06\ \text{m/s} \]

Figure 3.2 Example 3.1 in motion. The ledger in the panel is the horizontal equation itself: start from \(\colL{mv_1}\), add the impulse of the push \(\colP{I_P}\) and of friction \(\colF{I_f}\), each bar starting where the last one ended, and you land on \(\colL{mv_2}\), the momentum at that moment. Change the push, its angle or the friction. Push too gently and the crate does not move: static friction's impulse cancels the push's exactly.

Steepen the push in Figure 3.2. More of \(P\) goes into pressing the crate down and less into moving it, and the normal force, friction and the friction impulse all grow. With the starting values, past about \(38^\circ\) the crate does not move at all, however long you push.

Example 3.2 — Braking on a grade

A \(1500\ \text{kg}\) car traveling at \(25\ \text{m/s}\) down a \(5^\circ\) slope brakes with a constant total braking force of \(9\ \text{kN}\) up the slope. How long does it take to stop?

Show solution

Along the slope, positive downhill. The weight's component \(\colVg{1500(9.81)\sin 5^\circ = 1282\ \text{N}}\) acts downhill and the braking force uphill. The normal force is perpendicular to the slope and drops out of this direction.

\[ 1500(25) + \left(\colVg{1282} - 9000\right)t = 0 \quad\Rightarrow\quad t = \frac{37\,500}{7718} = 4.86\ \text{s} \]

On the level it would stop in \(37\,500/9000 = 4.17\ \text{s}\). The weight's impulse along the slope adds \(0.69\ \text{s}\).

Example 3.3 — Up an incline and back

A \(2\ \text{kg}\) block is launched at \(8\ \text{m/s}\) up a \(20^\circ\) incline. The coefficients of friction are \(\mu_s = 0.3\) and \(\mu_k = 0.25\). (a) When does it stop? (b) What is its velocity \(2\ \text{s}\) after the launch?

Show solution

Perpendicular to the incline nothing moves, so \(N = 2(9.81)\cos 20^\circ = 18.44\ \text{N}\) and \(\colF{F_f} = \mu_k N = 4.609\ \text{N}\). The weight's component along the incline is \(\colVg{2(9.81)\sin 20^\circ = 6.710\ \text{N}}\), down the slope.

(a) Going up, positive up the slope. Friction acts down the slope, with the weight:

\[ 2(8) \colVg{{} - 6.710\,t_1} \colF{{} - 4.609\,t_1} = 0 \quad\Rightarrow\quad t_1 = \frac{16}{11.32} = 1.413\ \text{s} \]

Does it stay? Holding it would take a friction force of \(6.710\ \text{N}\), more than \(\mu_s N = 5.531\ \text{N}\). So it slides back down.

(b) Coming down, positive down the slope. Friction has turned around: it now acts up the slope. Write a new equation for the new interval, from \(t_1\) to \(2\ \text{s}\):

\[ 0 + \colVg{6.710(2 - 1.413)} \colF{{} - 4.609(2 - 1.413)} = 2\,v_2 \quad\Rightarrow\quad v_2 = 0.616\ \text{m/s} \]

down the incline. One equation across the whole \(2\ \text{s}\) would be wrong: the friction impulse is not \(\mu_k N\) times \(2\ \text{s}\) in one direction, because friction reversed when the block did.

Bodies joined by a cord

For bodies joined by an inextensible cord over a pulley, write the principle for each body separately, each along its own direction of motion. The cord's tension pulls on both, so it appears in both equations; adding them eliminates it when the pulley is light and smooth, because the tension is then the same on both sides.

Figure 3.3 Block \(A\) on a rough table, joined by a cord over a light, smooth pulley to block \(B\), which hangs. Forces are drawn to scale for \(m_A = 10\ \text{kg}\), \(m_B = 5\ \text{kg}\) and \(\mu_k = 0.2\), while the blocks move, except friction, drawn \(2.5\) times longer so it is easy to see. The weights end at the centers of gravity and the table's push \(N_A\) ends on the bottom of \(A\); friction \(\colF{F_f}\) starts on the bottom of \(A\), where it acts. The cord's tension \(T\) pulls \(A\) toward the pulley and \(B\) upward, from where it is tied.

Example 3.4 — Two blocks and a pulley

In Figure 3.3, \(m_A = 10\ \text{kg}\), \(m_B = 5\ \text{kg}\) and the coefficient of kinetic friction under \(A\) is \(0.2\). The system is released from rest. Find the speed of the blocks \(2\ \text{s}\) later, and the tension in the cord.

Show solution

The cord is inextensible, so both blocks have the same speed \(v\). Under \(A\), \(N_A = 98.1\ \text{N}\) and \(\colF{F_f} = 0.2(98.1) = 19.62\ \text{N}\).

Block \(A\), positive toward the pulley: \(\quad 0 + T(2) \colF{{} - 19.62(2)} = 10\,v\)

Block \(B\), positive down: \(\quad 0 + \colVg{49.05(2)} - T(2) = 5\,v\)

Add them, and the tension's impulses cancel:

\[ \colVg{49.05(2)} \colF{{} - 19.62(2)} = 15\,v \quad\Rightarrow\quad v = 3.92\ \text{m/s} \]

Then from \(A\)'s equation, \(2T = 10(3.924) + 39.24\), so \(T = 39.2\ \text{N}\). It is less than \(B\)'s weight, \(49.05\ \text{N}\), as it must be for \(B\) to speed up as it falls.

Check your understanding

Key takeaways

\(m\vvec_1 + \sum\int\Fvec\,dt = m\vvec_2\): momentum plus the impulse of every force gives the new momentum. In a plane, two scalar equations.
Impulse and momentum diagrams are the free-body diagram with every force times its time, between the two momenta.
Find normal forces from the direction in which nothing moves, then friction from the normal force.
If friction changes direction, split the interval where the velocity passes through zero.
For bodies joined by a cord, write one equation per body; adding them eliminates the tension.
Next, Lesson 4 handles forces that change with time.