Momentum and collisions

Physics · Senior school physics · about 10 min

Momentum describes how much motion an object carries in a particular direction. During a short collision, the forces between the objects can be very large, yet their combined momentum stays constant if the external impulse is negligible. This makes it possible to calculate an unknown final velocity without knowing the complicated contact force.

The key idea

For a system with negligible external impulse, total momentum is conserved:

\sum mv\big|_{\mathrm{before}}=\sum mv\big|_{\mathrm{after}}

Here $m$ is mass in $\mathrm{kg}$ and $v$ is signed velocity in $\mathrm{m/s}$ . Momentum $p=mv$ has units $\mathrm{kg\,m/s}$ .

Two horizontal panels show the same frictionless straight track, with a thin right-pointing positive x-axis beneath both. In the upper panel, before contact, a smaller trolley labelled A on the left moves right with an arrow marked u_A, while a larger trolley labelled B on the right moves left with an arrow marked u_B. Their front faces are a short distance apart. In the lower panel, after contact, the trolleys are joined face to face and move together to the right with one arrow v. Use identical trolley sizes in the two panels and no numerical values. The positive axis is a reference direction, not another force arrow. — Choose one axis and conserve signed total momentum across the collision.

Try it yourself

Collide two pucks in two dimensions and check that total momentum is kept.

Direction and impulse

Momentum is a vector. In a one-dimensional problem, select a positive direction before doing any arithmetic. Motion in the opposite direction has negative velocity, even when the object is moving quickly. A change in momentum comes from an impulse:

J=F_{\mathrm{net,avg}}\Delta t=\Delta p

Impulse has units $\mathrm{N\,s}$ , equivalent to $\mathrm{kg\,m/s}$ . The equation refers to the net force on the chosen object. When a ball rebounds, its velocity reverses sign, so the momentum change can be larger than either its initial or final momentum magnitude. Do not replace velocity with speed.

Types of collision

For two objects moving along one line, conservation gives

m_1u_1+m_2u_2=m_1v_1+m_2v_2.

The letters $u$ and $v$ mean velocities before and after impact. In a perfectly inelastic collision, the objects stick and share one final velocity. In an elastic collision, total kinetic energy is conserved as well as momentum. In other collisions, some kinetic energy becomes sound, heat or deformation. Momentum conservation still applies to the complete isolated system, but kinetic energy conservation generally does not. Internal collision forces cancel when the two objects are treated as one system.

A reliable solving method

Sketch the objects before and after contact, mark a positive axis and assign a signed velocity to each object. Write the momentum equation before inserting values. If the bodies join, use a common final velocity and their combined mass. A negative result means movement opposite to your chosen axis; it does not mean an impossible velocity. When the question also asks about energy, calculate kinetic energy separately with $K=\tfrac12 mv^2$ . The square removes the velocity sign. Finally, check that total momentum has the same units and value on both sides.

Worked example: Two trolleys that lock together

On a straight, nearly frictionless track, trolley A of mass $1.5\,\mathrm{kg}$ moves right at $4.0\,\mathrm{m/s}$ . Trolley B of mass $2.5\,\mathrm{kg}$ moves left at $1.0\,\mathrm{m/s}$ . They collide and lock together. Find their common velocity and the kinetic energy lost in the collision.

A side view of two wheeled trolleys on a level track just before they collide. Place trolley A on the left and B on the right with a visible small gap. A is visibly shorter than B. Draw a right-pointing velocity arrow above A labelled +4.0 m/s, and a left-pointing velocity arrow above B labelled -1.0 m/s. Write 1.5 kg inside A and 2.5 kg inside B. Above the track, a small rightward reference arrow indicates +x. Below, include a simple outline of the joined trolleys after collision, with a rightward arrow labelled v = ?; do not give a numerical final velocity or energy loss. — The incoming velocities have opposite signs on a right-positive axis.

Show the solution

Step 1: Choose right as positive. The signed initial velocities are $u_A=+4.0\,\mathrm{m/s}$ and $u_B=-1.0\,\mathrm{m/s}$ .

Step 2: Conserve momentum. The joined trolleys have a common velocity $v$ :

(1.5)(4.0)+(2.5)(-1.0)&=(1.5+2.5)v\\ 6.0-2.5&=4.0v\\ v&=\frac{3.5}{4.0}=0.875\,\mathrm{m/s}. \end{aligned}$$ The positive sign means that the combined trolley moves right. **Step 3: Calculate kinetic energy before and after.**

K_i=\frac12(1.5)(4.0)^2+\frac12(2.5)(1.0)^2=13.25,\mathrm{J}.

K_f=\frac12(4.0)(0.875)^2=1.53125,\mathrm{J}.

K_{\mathrm{lost}}=13.25-1.53125=11.71875,\mathrm{J}\approx11.72,\mathrm{J}.

**Answer: $0.875\,\mathrm{m/s}$ right, with $11.72\,\mathrm{J}$ of kinetic energy lost.**

Common mistakes

Using positive speeds for both approaching objects

Choose an axis and make one initial velocity negative when it points opposite the axis.

Conserving kinetic energy when the objects stick

Conserve momentum, then calculate the change in kinetic energy separately.

Forgetting that impulse is a change, not a final momentum

Use $\Delta p=p_f-p_i$ with signed velocities and units of $\mathrm{N\,s}$ .

Treating a negative final velocity as an error

Interpret its sign as the direction of motion relative to the axis you selected.

Check your understanding

1. A $0.20\,\mathrm{kg}$ ball initially travels at $+8.0\,\mathrm{m/s}$ and rebounds at $-6.0\,\mathrm{m/s}$ . What is its change in momentum?

2. A $3.0\,\mathrm{kg}$ trolley moving at $+2.0\,\mathrm{m/s}$ sticks to a stationary $1.0\,\mathrm{kg}$ trolley. What is their common velocity?

3. For an isolated pair of objects, which quantities are conserved in an elastic collision?

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