Momentum and impulse in two dimensions

Key ideas

Momentum is a vector: $p = mv$ , pointing the same way as the velocity. Its unit is kg m s⁻¹.
Conservation of momentum still holds when there is no external force — but in two dimensions you must conserve it as a vector.
The reliable method is to split momentum into components:
- conserve the $x$ -component of total momentum: $\sum p_{x,\text{before}} = \sum p_{x,\text{after}}$
- conserve the $y$ -component of total momentum: $\sum p_{y,\text{before}} = \sum p_{y,\text{after}}$
Impulse is the change in momentum, and equals force × time:

$\text{impulse} = F\,\Delta t = \Delta p = m v_f - m v_i$

Impulse is also a vector — a sideways force changes the sideways momentum only.

Two objects colliding and sticking together: a momentum vector pointing east is added tip-to-tail to a momentum vector pointing north, and the resultant momentum arrow is the diagonal of the rectangle they form, labelled with its magnitude and angle.

Method for a 2D collision

Draw a vector diagram of the momenta before and after.
Resolve every momentum into $x$ - and $y$ -components.
Equate total before = total after, separately for $x$ and $y$ .
Recombine the final components with Pythagoras and trigonometry to get the final speed and direction.

Worked ExampleTwo cars stick together at an intersection

A $1200$ kg car moving east at $20$ m s⁻¹ collides with a $1500$ kg car moving north at $15$ m s⁻¹. They lock together. Find their common velocity just after impact.

Step 1 — Momentum components before (no external force during the impact)

East ( $x$ ): $p_x = (1200)(20) = 2.40 \times 10^4$ kg m s⁻¹ North ( $y$ ): $p_y = (1500)(15) = 2.25 \times 10^4$ kg m s⁻¹

Step 2 — Total momentum (Pythagoras)

$p = \sqrt{(2.40\times10^4)^2 + (2.25\times10^4)^2} = 3.29 \times 10^4 \text{ kg m s}^{-1}$

Step 3 — Common velocity (they share the total mass)

$v = \frac{p}{m_{total}} = \frac{3.29\times10^4}{1200 + 1500} = 12.2 \text{ m s}^{-1}$

Step 4 — Direction

$\theta = \tan^{-1}\!\left(\frac{p_y}{p_x}\right) = \tan^{-1}\!\left(\frac{2.25}{2.40}\right) = 43.2° \text{ north of east}$

Tips

Resolve into components before doing anything else. Trying to add momentum vectors by their magnitudes alone is the classic 2D-momentum mistake — a north momentum and an east momentum do not add to give their arithmetic sum.
Check your angle is measured from the axis you claim ("north of east" vs "east of north") — a protractor mark on your vector diagram earns method marks even if the arithmetic slips.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $0.50$ kg ball moving east at $4.0$ m s⁻¹ is struck and moves north at $3.0$ m s⁻¹.

Show that the magnitude of its change in momentum is $2.5$ kg m s⁻¹.

Merit

A $2.0$ kg trolley moving east at $3.0$ m s⁻¹ collides with a stationary $1.0$ kg trolley and they stick together.

Calculate their common speed.

Excellence

A $3.0$ kg object at rest explodes into two pieces. A $1.0$ kg piece flies east at $6.0$ m s⁻¹.

Determine the velocity of the $2.0$ kg piece, and explain why kinetic energy is not conserved here.

Momentum and impulse in two dimensions

Covers AS91524

Key ideas

Momentum is a vector: $p = mv$ , pointing the same way as the velocity. Its unit is kg m s⁻¹.
Conservation of momentum still holds when there is no external force — but in two dimensions you must conserve it as a vector.
The reliable method is to split momentum into components:
- conserve the $x$ -component of total momentum: $\sum p_{x,\text{before}} = \sum p_{x,\text{after}}$
- conserve the $y$ -component of total momentum: $\sum p_{y,\text{before}} = \sum p_{y,\text{after}}$
Impulse is the change in momentum, and equals force × time:

$\text{impulse} = F\,\Delta t = \Delta p = m v_f - m v_i$

Impulse is also a vector — a sideways force changes the sideways momentum only.

Method for a 2D collision

Draw a vector diagram of the momenta before and after.
Resolve every momentum into $x$ - and $y$ -components.
Equate total before = total after, separately for $x$ and $y$ .
Recombine the final components with Pythagoras and trigonometry to get the final speed and direction.

Worked ExampleTwo cars stick together at an intersection

A $1200$ kg car moving east at $20$ m s⁻¹ collides with a $1500$ kg car moving north at $15$ m s⁻¹. They lock together. Find their common velocity just after impact.

Step 1 — Momentum components before (no external force during the impact)

East ( $x$ ): $p_x = (1200)(20) = 2.40 \times 10^4$ kg m s⁻¹ North ( $y$ ): $p_y = (1500)(15) = 2.25 \times 10^4$ kg m s⁻¹

Step 2 — Total momentum (Pythagoras)

$p = \sqrt{(2.40\times10^4)^2 + (2.25\times10^4)^2} = 3.29 \times 10^4 \text{ kg m s}^{-1}$

Step 3 — Common velocity (they share the total mass)

$v = \frac{p}{m_{total}} = \frac{3.29\times10^4}{1200 + 1500} = 12.2 \text{ m s}^{-1}$

Step 4 — Direction

$\theta = \tan^{-1}\!\left(\frac{p_y}{p_x}\right) = \tan^{-1}\!\left(\frac{2.25}{2.40}\right) = 43.2° \text{ north of east}$

Tips

Resolve into components before doing anything else. Trying to add momentum vectors by their magnitudes alone is the classic 2D-momentum mistake — a north momentum and an east momentum do not add to give their arithmetic sum.
Check your angle is measured from the axis you claim ("north of east" vs "east of north") — a protractor mark on your vector diagram earns method marks even if the arithmetic slips.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $0.50$ kg ball moving east at $4.0$ m s⁻¹ is struck and moves north at $3.0$ m s⁻¹.

Show that the magnitude of its change in momentum is $2.5$ kg m s⁻¹.

Merit

A $2.0$ kg trolley moving east at $3.0$ m s⁻¹ collides with a stationary $1.0$ kg trolley and they stick together.

Calculate their common speed.

Excellence

A $3.0$ kg object at rest explodes into two pieces. A $1.0$ kg piece flies east at $6.0$ m s⁻¹.

Determine the velocity of the $2.0$ kg piece, and explain why kinetic energy is not conserved here.