Lesson 2 · 30 min
Linear Momentum and Impulse
The integrated form of Newton's second law has a name for each side. On the right, the change in a quantity that goes with the motion: linear momentum, mass times velocity. On the left, the total effect of the forces over the time they act: their impulse, which you can read off a force–time graph as an area.
Learning objectives
- Calculate the linear momentum \(\Lvec = m\vvec\) of a particle, as a vector and in components.
- Calculate the impulse of a constant force, \(\Fvec\,\Delta t\), and of a force that varies, \(\int\Fvec\,dt\).
- Read an impulse as the area under a force–time graph, and find the average force over an interval.
- Explain why every force acting during an interval has an impulse, including forces that do no work.
Linear momentum
The linear momentum of a particle of mass \(m\) moving with velocity \(\vvec\) is
Linear momentum
\[ \colL{\Lvec = m\vvec} \]A vector in the direction of the velocity, with magnitude \(mv\). Units: \(\text{kg·m/s}\).
- Momentum has a direction. In components, \(L_x = m v_x\) and \(L_y = m v_y\). A \(0.5\ \text{kg}\) ball moving at \(8\ \text{m/s}\), \(30^\circ\) above the horizontal, has \(L_x = 0.5(8\cos 30^\circ) = 3.46\ \text{kg·m/s}\) and \(L_y = 0.5(8\sin 30^\circ) = 2\ \text{kg·m/s}\).
- On a straight line, choose a positive direction and give each momentum a sign. A ball moving left at \(4\ \text{m/s}\) has \(mv = -4m\) when right is positive.
- Momentum grows in proportion to the speed; kinetic energy grows with its square. The table shows how differently the two rank the same objects.
| Object | Mass | Speed | Momentum \(mv\) | Kinetic energy \(\tfrac12 mv^2\) |
|---|---|---|---|---|
| Car | \(1500\ \text{kg}\) | \(20\ \text{m/s}\) | \(30\,000\ \text{kg·m/s}\) | \(300\ \text{kJ}\) |
| Rifle bullet | \(10\ \text{g}\) | \(800\ \text{m/s}\) | \(8\ \text{kg·m/s}\) | \(3.2\ \text{kJ}\) |
The car has \(3750\) times the bullet's momentum but only about \(94\) times its kinetic energy. To stop the car in the same time takes \(3750\) times the force.
The impulse of a force
The linear impulse of a force \(\Fvec\) acting from \(t_1\) to \(t_2\) is its integral over that time:
Linear impulse
\[ \Ivec = \int_{t_1}^{t_2} \Fvec\,dt \qquad\text{and, for a constant force,}\qquad \Ivec = \Fvec\,(t_2 - t_1) \]A vector. Units: \(\text{N·s}\), which is the same as \(\text{kg·m/s}\), the unit of momentum.
- A constant force's impulse points the same way as the force. In components, \(I_x = \int F_x\,dt\) and \(I_y = \int F_y\,dt\).
- Impulse needs only time, so every force acting during the interval has one, whether or not the body moves and whether or not the force does work. A \(10\ \text{kg}\) crate resting on the floor for \(5\ \text{s}\) receives an impulse of \(\colVg{98.1(5) = 490.5\ \text{N·s}}\) downward from its weight and the same, upward, from the floor. They cancel, so its momentum does not change.
- Compare work: the floor's normal force does no work on the crate, but it has an impulse. Work and impulse are different integrals of the same force.
Example 2.1 — The impulses on a pulled crate
A \(20\ \text{kg}\) crate on a smooth floor is pulled for \(3\ \text{s}\) by a rope with a constant tension of \(100\ \text{N}\), at \(30^\circ\) above the horizontal. Find the impulse of each force on the crate over the \(3\ \text{s}\), and the crate's speed at the end if it starts from rest.
Show solution
The rope. \(\colP{I_x = 100\cos 30^\circ\,(3) = 259.8\ \text{N·s}}\) and \(\colP{I_y = 100\sin 30^\circ\,(3) = 150\ \text{N·s}}\).
The weight. \(W = 20(9.81) = 196.2\ \text{N}\), so \(\colVg{I = 196.2(3) = 588.6\ \text{N·s}}\), downward.
The floor. The crate stays on the floor, so the vertical forces balance: \(N = 196.2 - 50 = 146.2\ \text{N}\), and its impulse is \(146.2(3) = 438.6\ \text{N·s}\), upward. The vertical impulses add to zero: \(150 + 438.6 - 588.6 = 0\).
The speed. Only the rope has a horizontal impulse, and it equals the crate's change in momentum:
\[ 259.8 = 20\,v_2 - 0 \quad\Rightarrow\quad v_2 = 13.0\ \text{m/s} \]Three forces, three impulses, and two of them belong to forces that do no work.
Impulse as an area
On a graph of force against time, \(\int F\,dt\) is the area between the curve and the time axis: areas above the axis count as positive impulse, areas below as negative. The average force over an interval is the constant force that would give the same area over the same time:
\[ F_{\text{avg}} = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2} F\,dt \]Try the shapes with the same peak and duration. The rectangle gives the most impulse; the triangle gives half as much; the half-sine gives \(2/\pi\), about \(64\%\). The reversing force pushes, then pulls just as hard: its two areas cancel, and the block ends at rest.
Example 2.2 — Reading a force–time graph
A \(10\ \text{kg}\) block rests on a smooth horizontal surface. A horizontal force increases uniformly from \(0\) to \(30\ \text{N}\) in \(2\ \text{s}\), stays at \(30\ \text{N}\) for \(2\ \text{s}\), then decreases uniformly to zero at \(t = 5\ \text{s}\). Find the impulse of the force, the average force, and the block's speed at \(t = 5\ \text{s}\).
Show solution
The area is a triangle, a rectangle and a triangle:
\[ I = \tfrac12(2)(30) + (2)(30) + \tfrac12(1)(30) = 30 + 60 + 15 = 105\ \text{N·s} \]The average force is \(F_{\text{avg}} = 105/5 = 21\ \text{N}\). Only this force acts horizontally, so its impulse is the block's change in momentum:
\[ 0 + 105 = 10\,v_2 \quad\Rightarrow\quad v_2 = 10.5\ \text{m/s} \]Example 2.3 — A baseball and a bat
A \(0.145\ \text{kg}\) baseball arrives horizontally at \(40\ \text{m/s}\), and the bat sends it straight back at \(50\ \text{m/s}\). The contact lasts \(1.2\ \text{ms}\). Find the impulse of the bat on the ball and the average force.
Show solution
Take the direction of the hit as positive. The ball arrives with \(v_1 = -40\ \text{m/s}\) and leaves with \(v_2 = +50\ \text{m/s}\). Over the very short contact, the bat's force is the only one that matters:
\[ \colI{I} = m v_2 - m v_1 = 0.145\left[50 - (-40)\right] = 13.05\ \text{N·s} \] \[ \colI{F_{\text{avg}}} = \frac{I}{\Delta t} = \frac{13.05}{0.0012} = 10\,900\ \text{N} = 10.9\ \text{kN} \]The ball's weight, \(1.42\ \text{N}\), is about \(7600\) times smaller, and its impulse during the contact, \(0.0017\ \text{N·s}\), is negligible. A force like the bat's, very large and very brief, is called impulsive; Lesson 5 uses the distinction.
The most common slip is writing the change in velocity as \(50 - 40\). Velocities are vectors: the rebound reverses the sign, so the change is \(50 - (-40) = 90\ \text{m/s}\).
Check your understanding
Key takeaways
- Linear momentum \(\Lvec = m\vvec\) is a vector, in \(\text{kg·m/s}\). On a line, give each momentum a sign.
- The impulse of a force is \(\int\Fvec\,dt\), or \(\Fvec\,\Delta t\) if the force is constant, in \(\text{N·s} = \text{kg·m/s}\).
- Impulse is the area under the force–time graph; the average force is that area divided by the time.
- Every force acting during the interval has an impulse, including the weight and normal forces that do no work.
- Next, Lesson 3 puts momentum and impulse together in the principle of linear impulse and momentum.