3.4 Momentum — Learn

Prerequisites: Newton's laws and $F = ma$ from Section 3.2; velocity and the sign convention for one-dimensional motion from Section 3.1; kinetic energy from Section 3.3.

By the end of this section you can: define momentum; use the impulse–momentum relation $F\Delta t = \Delta p$ ; read a force–time graph; and apply conservation of momentum to one-dimensional collisions, explosions and recoil.

1. Momentum

The momentum of an object is its mass multiplied by its velocity.

p = mv

Momentum is a vector, so it has the same direction as the velocity. Its unit is the kilogram metre per second, $\text{kg}\,\text{m}\,\text{s}^{-1}$ , which is also written as the newton second, $\text{N}\,\text{s}$ .

A larger mass or a larger velocity gives more momentum. A heavy lorry moving slowly can have the same momentum as a small car moving quickly. Stopping the lorry takes the same total change in momentum as stopping the car.

For motion along a line, choose one direction as positive. An object moving the other way has negative momentum. Adding the signed momenta of a system gives the total momentum, and its sign gives the direction of the total.

Worked example 1. A car of mass 1200 kg travels at $15\,\text{m}\,\text{s}^{-1}$ due east. Find its momentum.

p = mv = 1200 \times 15 = 1.8 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1} \ \text{east}

2. Newton's second law and impulse

Newton's second law can be written in terms of momentum. If the mass is constant,

F = \frac{\Delta p}{\Delta t}

The resultant force equals the rate of change of momentum. For a force that acts for a time $\Delta t$ , rearranging gives the impulse:

F\Delta t = \Delta p

Impulse is measured in newton seconds. The impulse on an object equals its change in momentum. On a force–time graph, the area under the graph is the impulse, and the change in momentum.

This form of the law explains why a collision is safer when it lasts longer. For a fixed change in momentum, a longer contact time gives a smaller force. Crumple zones and airbags make a crash last longer, so the force on the passengers is smaller.

Worked example 2. A ball of mass 0.20 kg hits a wall at $8.0\,\text{m}\,\text{s}^{-1}$ and rebounds at $6.0\,\text{m}\,\text{s}^{-1}$ . It is in contact with the wall for 0.050 s. Find the change in momentum and the average force on the ball.

Take the direction towards the wall as positive, so $u = +8.0$ and $v = -6.0$ .

\begin{aligned} \Delta p &= m(v - u) \\ &= 0.20(-6.0 - 8.0) \\ &= -2.8\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ F &= \frac{\Delta p}{\Delta t} = \frac{-2.8}{0.050} \\ &= -56\,\text{N} \end{aligned}

The force is 56 N away from the wall. The negative sign shows that the force on the ball points in the negative direction, away from the wall.

Worked example 3. A car of mass 900 kg is brought to rest from $20\,\text{m}\,\text{s}^{-1}$ in 4.0 s. Find the average braking force.

\begin{aligned} \Delta p &= 0 - 900 \times 20 = -18000\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ F &= \frac{\Delta p}{\Delta t} = \frac{-18000}{4.0} = -4500\,\text{N} \end{aligned}

The braking force is 4500 N opposite to the motion.

3. Conservation of momentum

If no resultant external force acts on a system, its total momentum is conserved.

\sum p_{\text{before}} = \sum p_{\text{after}}

This applies to collisions, to explosions and to recoil. The forces between the objects are internal, and by Newton's third law they are equal and opposite, so they change the two momenta by equal and opposite amounts. The total is unchanged.

Worked example 4. Trolley A of mass 2.0 kg moves at $3.0\,\text{m}\,\text{s}^{-1}$ towards trolley B of mass 4.0 kg, which is at rest. The trolleys collide and stick together. Find their common velocity after the collision.

Take the initial direction of A as positive. Before the collision,

p_{\text{before}} = 2.0 \times 3.0 + 4.0 \times 0 = 6.0\,\text{kg}\,\text{m}\,\text{s}^{-1}

After the collision the total mass is 6.0 kg and the common velocity is $v$ .

\begin{aligned} 6.0 &= 6.0v \\ v &= 1.0\,\text{m}\,\text{s}^{-1} \end{aligned}

The trolleys move together at $1.0\,\text{m}\,\text{s}^{-1}$ in the original direction of A.

Worked example 5. A rifle of mass 4.0 kg fires a bullet of mass 0.020 kg at $400\,\text{m}\,\text{s}^{-1}$ . Find the recoil velocity of the rifle.

The rifle and bullet are at rest before firing, so the total momentum is zero.

\begin{aligned} 0 &= 0.020 \times 400 + 4.0v \\ v &= -\frac{8.0}{4.0} = -2.0\,\text{m}\,\text{s}^{-1} \end{aligned}

The rifle recoils at $2.0\,\text{m}\,\text{s}^{-1}$ in the opposite direction to the bullet.

4. Try it: the total momentum before and after

The figure shows two trolleys on a line. It stacks their momenta end to end, so you can see the total before the collision and after it. Four sliders set the two masses and the two velocities, and a menu chooses whether the trolleys stick together or bounce apart.

Predict first. Do not move the controls yet. Trolley A is $2\,\text{kg}$ and moves at $3\,\text{m}\,\text{s}^{-1}$ to the right. Trolley B is $4\,\text{kg}$ and is at rest. The trolleys stick together. Write down your answers.

Find the total momentum before the collision.
Is the total momentum after the collision larger, smaller or the same?
The trolleys move together after the collision. Do they move to the right, to the left, or not at all?

Test. Set the four controls to your prediction, and read the two totals. Check each value against Worked example 4. Now change one control at a time and watch the two totals. Then choose the option where the trolleys bounce apart, and check the two totals again.

Explain. Why is the total momentum the same before and after, when the velocity of each trolley changes? Write two or three sentences. Use Newton's third law and the impulse–momentum relation.

5. Collisions and kinetic energy

Momentum is conserved in every collision, but kinetic energy is not always conserved.

In an elastic collision, kinetic energy is conserved as well as momentum. A collision between gas molecules is close to elastic.
In an inelastic collision, some kinetic energy becomes thermal energy or sound. Momentum is still conserved.
In a perfectly inelastic collision, the objects stick together and move as one. This loses the most kinetic energy of any collision with the same momenta.

Worked example 6. In Worked example 4, check whether the collision is elastic.

\begin{aligned} E_k(\text{before}) &= \tfrac{1}{2} \times 2.0 \times 3.0^2 = 9.0\,\text{J} \\ E_k(\text{after}) &= \tfrac{1}{2} \times 6.0 \times 1.0^2 = 3.0\,\text{J} \end{aligned}

Kinetic energy has fallen from 9.0 J to 3.0 J, so the collision is inelastic. The missing 6.0 J has been transferred to thermal stores and sound.

Watch out. Momentum is a vector, so the direction signs matter. When an object bounces back, its velocity reverses, and the change in momentum is the final momentum minus the initial momentum. If you add the magnitudes instead of subtracting, you find the change as if the object had stopped. Also remember that momentum is conserved only when the resultant external force is zero; friction from the ground or a push from outside breaks the conservation.

Progress check

Return to the chapter problem. The car and riders have a mass of 500 kg and reach a speed of $29.7\,\text{m}\,\text{s}^{-1}$ at the bottom. At the end of the ride the brakes bring the car to rest in 3.0 s.

Find the momentum of the car at the bottom.
Find the change in momentum as the car stops.
Use $F\Delta t = \Delta p$ to find the average braking force.

If question 1 goes wrong, use $p = mv$ . If question 2 goes wrong, the final momentum is zero. If question 3 goes wrong, divide the change in momentum by the contact time of 3.0 s.

You should now be able to

define momentum and state that it is a vector;
use $p = mv$ with a sign convention along a line;
write Newton's second law as $F = \dfrac{\Delta p}{\Delta t}$ ;
use the impulse–momentum relation $F\Delta t = \Delta p$ ;
find the impulse from the area under a force–time graph;
explain why a longer collision time gives a smaller force;
apply conservation of momentum to a one-dimensional collision, an explosion and a recoil;
tell an elastic collision from an inelastic one using kinetic energy.