Physics 9702 · Paper 2

Momentum in Two Dimensions

Real collisions are rarely head-on. A ball hits another off centre, and the two move apart at angles.
Momentum is a vector, so it is conserved separately in each direction. That one sentence is the whole method.

Split the momentum into components

Choose two directions at right angles: $x$ along the original motion, $y$ across it.
Resolve each momentum with the skill from chapter 1: $p\cos\theta$ along, $p\sin\theta$ across.
Before the collision nothing moves across the line, so the total $y$ -momentum is zero. After the collision the two sideways components must cancel.
Along the line, the totals before and after are equal, as always.

An off-centre collision. Across the original line the two sideways momentum components must cancel, because there was no sideways momentum before.

Worked example

Smallest case: a symmetric split

A 2.0 kg ball moving at 6.0 m s⁻¹ hits an identical stationary ball. Both move off at 30° to the original line, at the same speed v. Find v.

Across the line: the two $\sin$ components are equal and opposite, so they cancel. Nothing to solve.
Along the line: $2.0 \times 6.0 = 2.0\,v\cos 30^\circ + 2.0\,v\cos 30^\circ$ .
$12 = 4.0\,v\cos 30^\circ$ , so $v = \dfrac{3.0}{\cos 30^\circ} = 3.5 \text{ m s}^{-1}$ .

Answer

v = 3.5 \text{ m s}^{-1}

Worked example

Exam version: different masses, both angles known

An object of mass 2m travels at 5.0 m s⁻¹ and hits a stationary object of mass 3m. Afterwards the 2m object moves at speed v and the 3m object at speed w, each at 30° on opposite sides of the original line. Find v and w. (9702/21/M/J/24 Q3(b))

Across the line (totals cancel): $2m\,v\sin 30^\circ = 3m\,w\sin 30^\circ$ , so $2v = 3w$ .
Along the line: $2m(5.0) = 2m\,v\cos 30^\circ + 3m\,w\cos 30^\circ$ , so $10 = (2v + 3w)\cos 30^\circ$ .
Substitute $3w = 2v$ : $10 = 4v\cos 30^\circ$ , so $v = 2.9 \text{ m s}^{-1}$ and $w = \tfrac{2}{3}v = 1.9 \text{ m s}^{-1}$ .

Answer

v = 2.9 \text{ m s}^{-1}, \; w = 1.9 \text{ m s}^{-1}

The vector triangle

There is a second picture of the same fact. Draw the momentum vectors after the collision tip to tail. Because momentum is conserved, they must add up to the momentum before.
So the three vectors form a closed triangle. If the triangle closes, momentum is conserved.

Momentum is conserved, so the two after-vectors add up to the before-vector: the triangle closes. Here the sides are a real 3-4-5 right triangle.

Multiple-choice questions use this picture: two momenta at right angles stick together, and the answer is the vector sum of the triangle (9702/12/O/N/25 Q10).
Harder variant: a nucleus decays while already moving, and you balance the sideways components of the two fragments against each other (9702/22/O/N/24 Q6(c)).

A well-known special case: a perfectly elastic collision between equal masses, one at rest, always sends the two off at 90° to each other.

PreviousExplosions & Recoil NextForce = Rate of Change of Momentum

Momentum in Two Dimensions

Real collisions are rarely head-on. A ball hits another off centre, and the two move apart at angles.
Momentum is a vector, so it is conserved separately in each direction. That one sentence is the whole method.

Split the momentum into components

Choose two directions at right angles: $x$ along the original motion, $y$ across it.
Resolve each momentum with the skill from chapter 1: $p\cos\theta$ along, $p\sin\theta$ across.
Before the collision nothing moves across the line, so the total $y$ -momentum is zero. After the collision the two sideways components must cancel.
Along the line, the totals before and after are equal, as always.

An off-centre collision. Across the original line the two sideways momentum components must cancel, because there was no sideways momentum before.

Worked example

Smallest case: a symmetric split

A 2.0 kg ball moving at 6.0 m s⁻¹ hits an identical stationary ball. Both move off at 30° to the original line, at the same speed v. Find v.

Across the line: the two $\sin$ components are equal and opposite, so they cancel. Nothing to solve.
Along the line: $2.0 \times 6.0 = 2.0\,v\cos 30^\circ + 2.0\,v\cos 30^\circ$ .
$12 = 4.0\,v\cos 30^\circ$ , so $v = \dfrac{3.0}{\cos 30^\circ} = 3.5 \text{ m s}^{-1}$ .

Answer

v = 3.5 \text{ m s}^{-1}

Worked example

Exam version: different masses, both angles known

Across the line (totals cancel): $2m\,v\sin 30^\circ = 3m\,w\sin 30^\circ$ , so $2v = 3w$ .
Along the line: $2m(5.0) = 2m\,v\cos 30^\circ + 3m\,w\cos 30^\circ$ , so $10 = (2v + 3w)\cos 30^\circ$ .
Substitute $3w = 2v$ : $10 = 4v\cos 30^\circ$ , so $v = 2.9 \text{ m s}^{-1}$ and $w = \tfrac{2}{3}v = 1.9 \text{ m s}^{-1}$ .

Answer

v = 2.9 \text{ m s}^{-1}, \; w = 1.9 \text{ m s}^{-1}

The vector triangle

There is a second picture of the same fact. Draw the momentum vectors after the collision tip to tail. Because momentum is conserved, they must add up to the momentum before.
So the three vectors form a closed triangle. If the triangle closes, momentum is conserved.

Momentum is conserved, so the two after-vectors add up to the before-vector: the triangle closes. Here the sides are a real 3-4-5 right triangle.

Multiple-choice questions use this picture: two momenta at right angles stick together, and the answer is the vector sum of the triangle (9702/12/O/N/25 Q10).
Harder variant: a nucleus decays while already moving, and you balance the sideways components of the two fragments against each other (9702/22/O/N/24 Q6(c)).

A well-known special case: a perfectly elastic collision between equal masses, one at rest, always sends the two off at 90° to each other.