Momentum, impulse and collisions

Until this week you always examined the motion of a single object. Momentum (linear momentum) is the most powerful tool for situations in which two or more objects interact. In a collision, the forces change in a very short time, very large and in a complicated way. Without knowing them you can still find what happens before and after the collision. Because if there is no external force, the total momentum does not change.

Golden rules:

1. $\vec p = m\vec v$. It is a vector, so sign and component matter.
2. If there is no external force the total momentum is conserved. In all collisions (elastic or not).
3. Kinetic energy is conserved only in an elastic collision.
4. In two dimensions the x and y components are conserved separately.

Linear momentum

The momentum of an object is the product of its mass and its velocity:

$\vec p = m\vec v \qquad [p] = \text{kg}\cdot\text{m/s}$

Newton actually wrote the second law with momentum:

$\sum\vec F = \frac{d\vec p}{dt}$

If the mass is constant, this equals $m\vec a$. But the momentum form is more general; it holds in systems with changing mass, like a rocket.

Conservation of momentum

When two objects collide, the force 1 exerts on 2 is equal and opposite to the force 2 exerts on 1 (Newton 3). So the changes in momentum are also equal and opposite, and their sum is zero. A system with no external force (or where it is neglected) is called an isolated system:

$\vec p_{1i} + \vec p_{2i} = \vec p_{1f} + \vec p_{2f}$

TIP · why is gravity neglected in a collision? Collision forces are hundreds of times larger than the weight, and the collision lasts a very short time. The impulse of the weight in that short time is small enough to be neglected.

Impulse

The effect of a force over a period of time is called impulse. The impulse-momentum theorem:

$\vec J = \int \vec F\,dt = \Delta\vec p = m\vec v_f - m\vec v_i$

$\vec J = \vec F_{avg}\,\Delta t \qquad [J] = \text{N·s} = \text{kg·m/s}$

Impulse is the area under the F-t graph. If you make the same change in momentum over a longer time, the force is smaller. The airbag, landing by bending the knees and the helmet are all based on this idea.

Worked example 1. A 0.160 kg ball is met by a racket: it arrives at 20.0 m/s and goes off in the opposite direction at 25.0 m/s. The contact is 4.00 ms. What are the impulse and the average force?

Direction of the ball's departure +:  v_i = -20.0 m/s,  v_f = +25.0 m/s
J = m (v_f - v_i) = 0.160 · (25.0 - (-20.0)) = 0.160 · 45.0 = 7.20 N·s
F_avg = J / Δt = 7.20 / 0.00400 = 1.80 × 10³ N

PITFALL · in a rebound the speeds add. For an object that reverses direction, the change in velocity is not $25 - 20 = 5$ but $25 - (-20) = 45$ m/s. Whoever writes it without signs finds an impulse 9 times too small.

Types of collisions

The lost kinetic energy goes into heat, sound and deformation. The total energy is still conserved; only the mechanical energy is not.

Perfectly inelastic collision

The objects stick and move together:

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$

Worked example 2 (ballistic pendulum). A 10.0 g bullet at 400 m/s embeds itself in a 2.00 kg wooden block that hangs from a string. How high does the block rise?

Collision (momentum):  0.0100 · 400 = 2.01 · v   →  v = 1.99 m/s
Kinetic energy: before 800 J, after ½ · 2.01 · 1.99² = 3.98 J (most became heat)
Rise (energy, AFTER the collision):  v² = 2 g h  →  h = 1.99² / 19.6 = 0.202 m

Separate the two stages. At the moment of the collision, energy is not conserved, only momentum. In the swing that follows, mechanical energy is conserved.

Elastic collision in one dimension

Momentum and kinetic energy are conserved together. If $m_2$ is at rest:

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i} \qquad v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i}$

In the general case, use this linear relation instead of the energy equation with squares: $v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$. The speed of approach equals the speed of separation.

Worked example 3. A 2.00 kg ball hits a stationary 3.00 kg ball elastically at 5.00 m/s.

v1f = (2 - 3)/(2 + 3) · 5.00 = -1.00 m/s   (the light one bounced back)
v2f = 2·2/(2 + 3) · 5.00 = +4.00 m/s
Check, K: ½·2·1² + ½·3·4² = 1 + 24 = 25 J = ½·2·5²  ✓

Special cases: with equal masses the velocities are exchanged. An object hitting a very heavy target ($m_2 \gg m_1$) bounces back at the same speed. A heavy object hitting a very light target hardly slows down, and the light one shoots off at $2v$.

Collisions in two dimensions

Since momentum is a vector, the x and y components are conserved separately:

$\sum p_{x,i} = \sum p_{x,f} \qquad \sum p_{y,i} = \sum p_{y,f}$

Worked example 4. A 0.300 kg disc moving at 4.00 m/s in the $+x$ direction hits a stationary identical disc with a glancing collision. The first disc leaves at $37.0^\circ$ above $+x$ at 3.20 m/s. What is the velocity of the second disc?

x:  0.300·4.00 = 0.300·3.20·cos37 + 0.300·v2x  →  v2x = 4.00 - 2.556 = 1.44 m/s
y:  0 = 0.300·3.20·sin37 + 0.300·v2y           →  v2y = -1.93 m/s
v2 = √(1.44² + 1.93²) = 2.41 m/s,  angle = atan(-1.93/1.44) = -53.1° (below x)

The angle between the directions of the two discs is $37.0 + 53.1 = 90^\circ$. Equal masses always separate at a right angle in an elastic glancing collision.

PITFALL · finding the angle from the speeds. For two cars that stick together, the direction is $\tan^{-1}(p_y/p_x)$. If the masses are different, $\tan^{-1}(v_y/v_x)$ gives the wrong angle.

Is the collision elastic? Test it

On the exam, sometimes the speeds before and after the collision are given and you are asked “is it elastic”. First check the momentum (it must always be conserved), then compare the kinetic energy.

Worked example 5. A 0.500 kg ball hits a stationary 1.00 kg ball at 4.00 m/s. Afterward the light ball goes at $-1.00$ m/s and the heavy ball at $+2.50$ m/s.

Momentum before: 0.500 · 4.00               = 2.00 kg·m/s
Momentum after:  0.500·(-1.00) + 1.00·2.50  = 2.00 kg·m/s   ✓ conserved
K before: ½ · 0.500 · 4.00²                 = 4.00 J
K after:  ½·0.500·1.00² + ½·1.00·2.50²      = 0.250 + 3.125 = 3.375 J
Loss: 0.625 J  →  an inelastic collision

If momentum is not conserved, the speeds are physically impossible or there is an external force. If the kinetic energy has increased, there is a mistake too (except an explosion, where internal energy is converted to kinetic).

Which equation should I use?

Common mistakes