C. Energy and Momentum · Momentum and Collisions

Momentum in Two Dimensions

C2.5 C2.6 A1.12About 70 minutes

Learning goals

I can resolve momentum vectors into components and add them.
I can apply conservation of momentum separately to x- and y-components.
I can find the change in momentum of an object that changes direction in two dimensions.
I can solve two-dimensional explosion and sticking-collision problems.

Words to know

component
The part of a vector along one axis, such as p_x = p cos θ.

resultant
The single vector equal to the sum of several vectors.

tip-to-tail method
Adding vectors by drawing each one starting at the tip of the one before; the resultant runs from the first tail to the last tip.

direction notation [E 30° N]
A direction found by starting at east and turning 30° toward north.

Off the line

Real collisions rarely happen along one straight line. Two cars meet at an intersection. A puck glances off another puck or bounces off the boards. A firework bursts in every direction. To handle these, you need momentum as a vector in two dimensions.

Remember from Grade 11: in the SPH3U Kinematics: Motion in One and Two Dimensions unit you resolved vectors into components and added them, and you wrote directions like [E 30° N] (start facing east, turn 30° toward north). In the Dynamics: Forces and Motion in a Plane unit of this course you resolved forces in two dimensions. Momentum works exactly the same way.

Momentum vectors and components

Momentum p = mv points in the direction of the velocity. In two dimensions, resolve each momentum into x (east-west) and y (north-south) components:

p_x = p cos θ and p_y = p sin θ, where θ is measured from the +x axis (east). Use signs: west and south components are negative.

Total momentum: add all the x-components and all the y-components separately:

p_x(total) = p₁ₓ + p₂ₓ + ... and p_y(total) = p₁ᵧ + p₂ᵧ + ...

Then the size is p = √(p_x² + p_y²) and the direction is θ = tan⁻¹(|p_y| ÷ |p_x|), measured from the x-axis toward the y-axis, with the quadrant read from the signs.

Conservation of momentum in two dimensions

If the net external force on a system is zero, its total momentum vector is constant. A vector is constant only if each of its components is constant. So in two dimensions you get two equations:

Σp_x (before) = Σp_x (after) and Σp_y (before) = Σp_y (after)

Two equations means you can solve for two unknowns: for example, the speed and direction of one object, or two speeds when the directions are known.

Worked example

Example 1: Adding two momenta

Find the total momentum of a 2.0 kg object moving at 3.0 m/s [E] and a 1.5 kg object moving at 4.0 m/s [N].

p₁ = (2.0)(3.0) = 6.0 kg·m/s [E]: components (6.0, 0).
p₂ = (1.5)(4.0) = 6.0 kg·m/s [N]: components (0, 6.0).
Total: (6.0, 6.0). Size: √(6.0² + 6.0²) ≈ 8.5 kg·m/s. Direction: tan⁻¹(6.0 ÷ 6.0) = 45°.

8.5 kg·m/s [E 45° N]. Notice that 6.0 + 6.0 = 12 is wrong: perpendicular vectors do not add like numbers.

Worked example

Example 2: A collision at an intersection

A 1500 kg car moving at 15 m/s [E] collides with a 2000 kg SUV moving at 12 m/s [N]. They lock together. Find the velocity of the wreck just after the collision and the percentage of kinetic energy lost.

x: (1500)(15) + 0 = 22 500 kg·m/s. y: 0 + (2000)(12) = 24 000 kg·m/s.
Total p = √(22 500² + 24 000²) ≈ 32 900 kg·m/s. Direction: tan⁻¹(24 000 ÷ 22 500) ≈ 46.8° north of east.
v′ = 32 900 ÷ 3500 ≈ 9.40 m/s.
Ek before = ½(1500)(15)² + ½(2000)(12)² = 168 750 + 144 000 = 312 750 J. Ek after = ½(3500)(9.40)² ≈ 154 600 J.
Lost: (312 750 − 154 600) ÷ 312 750 ≈ 0.51.

The wreck moves at about 9.4 m/s [E 46.8° N], and about 51% of the kinetic energy is lost.

Example 2 drawn as a vector diagram. The car's eastward momentum and the SUV's northward momentum, placed tip to tail, add to the total momentum of the wreck: about 32 900 kg·m/s at 46.8° north of east.

Worked example

Example 3: Change in momentum off the boards

A 0.16 kg puck slides at 20 m/s toward the boards, making a 40° angle with them. It bounces off at the same speed and the same 40° angle. Find its change in momentum. If the puck touches the boards for 0.010 s, find the average force of the boards on it.

Let x be along the boards and y be perpendicular to them, pointing away from the boards.
Before: p_x = (0.16)(20)cos 40° ≈ 2.45; p_y = −(0.16)(20)sin 40° ≈ −2.06 kg·m/s.
After: p_x ≈ 2.45 (unchanged); p_y ≈ +2.06 kg·m/s.
Δp_x = 0 and Δp_y = 2.06 − (−2.06) ≈ 4.11 kg·m/s.
F_avg = Δp ÷ Δt = 4.11 ÷ 0.010 ≈ 411 N.

Δp ≈ 4.1 kg·m/s, perpendicular to the boards, pointing away from them. The boards push on the puck with about 410 N in that direction. The component along the boards does not change, because the (smooth) boards can only push perpendicular to their surface.

Left: the puck's path off the boards. Right: the vector triangle. The change in momentum Δp is the vector you add to the initial momentum to get the final momentum. It points straight away from the boards. (Arrows to scale.)

Worked example

Example 4: An explosion in two dimensions

An object at rest explodes into three pieces. A 1.0 kg piece moves at 6.0 m/s [E]. A 2.0 kg piece moves at 4.0 m/s [N]. The third piece has a mass of 1.5 kg. Find its velocity.

Initial momentum is zero, so the three final momenta must add to zero.
p₁ = (6.0, 0) and p₂ = (0, 8.0) kg·m/s.
p₃ = −(p₁ + p₂) = (−6.0, −8.0) kg·m/s. Its size is √(6.0² + 8.0²) = 10 kg·m/s.
Direction: tan⁻¹(8.0 ÷ 6.0) ≈ 53.1° south of west.
v₃ = 10 ÷ 1.5 ≈ 6.7 m/s.

The third piece moves at about 6.7 m/s [W 53.1° S].

The momenta in Example 4 drawn tip to tail. Because the total momentum is zero, the arrows return to their starting point and form a closed triangle.

Two ways to add vectors

Components (shown above) work for any number of vectors and any angles. For two vectors, you can also draw them tip to tail and use the cosine law and sine law from MCR3U. A careful scale diagram with a ruler and protractor is a good check on either method.

Just for fun

I tried to tell a joke about momentum in two dimensions, but it went off on a tangent.

Luckily, its x-component and its y-component were both conserved.

Hands-on activity

Draw a momentum map

You need: a large sheet of paper (or graph paper), a ruler, a protractor and a sharp pencil.

Choose a scale, such as 1 cm = 1 kg·m/s. Mark a north arrow.
Draw the momenta of Example 4 tip to tail: 6.0 cm east, then 8.0 cm north. Close the triangle by drawing the third side back to the start. Measure its length and its angle with the protractor. Compare with 10 kg·m/s at [W 53.1° S].
Now invent your own explosion. Draw any three arrows tip to tail that close into a triangle. Measure each arrow and its direction. Give each piece a mass and calculate its speed.
Swap with someone at home. Give them the masses and velocities of two of your pieces and the mass of the third, and have them calculate the third velocity using components. Check against your drawing.

What to look for: measured lengths within about 2 mm and angles within about 2° of the calculated values. If the triangle does not close, the total momentum is not zero, and the explosion could not have started from rest.

Your turn

Find the total momentum of a 3.0 kg object moving at 2.0 m/s [N] and a 2.0 kg object moving at 4.0 m/s [W].

Find the x- and y-components of a momentum of 12 kg·m/s [E 30° N].

An object at rest explodes into two pieces. A 1.0 kg piece moves at 6.0 m/s [E 20° N]. Find the velocity of the 3.0 kg piece.

A 0.20 kg puck moving at 3.0 m/s [E] and a 0.30 kg puck moving at 2.0 m/s [N] collide and stick. Find their velocity.

A 0.40 kg ball hits a wall at 5.0 m/s, making a 30° angle with the wall, and bounces off at the same speed and angle. Find its change in momentum.

An object at rest explodes into three pieces: 2.0 kg at 3.0 m/s [N], 1.0 kg at 4.0 m/s [E], and a third piece of 1.0 kg. Find the third piece's velocity.

More practice: warm-up

In a two-dimensional collision of an isolated system, what exactly is conserved?

Give the components of 10 kg·m/s [N] (x east, y north).

Find the size of a momentum with components (3.0, 4.0) kg·m/s.

Two momenta, each of size p, are perpendicular. What is the size of their sum?

What does the direction [E 30° N] mean?

More practice: on your own

A 1200 kg car moving at 20 m/s [E] collides with an 1800 kg truck moving at 10 m/s [S]. They lock together. Find their velocity and the percentage of kinetic energy lost.

A 2.0 kg object moving at 3.0 m/s [E] splits into two 1.0 kg pieces. One moves at 4.0 m/s [N]. Find the velocity of the other.

A stick deflects a 0.16 kg puck from 10 m/s [E] to 10 m/s [N] in 0.010 s. Find the change in momentum and the average force on the puck.

A 60 kg skater moving at 4.0 m/s [N] collides with an 80 kg skater moving at 3.0 m/s [E], and they hold on. Find their velocity.

Use the cosine law to add momenta of 8.0 kg·m/s [E] and 6.0 kg·m/s [E 60° N]. Check with components.

Challenge

A shell at rest at the top of its flight bursts into three equal fragments. Two of them fly off horizontally at 20 m/s, at 120° to each other. Show that the third also moves at 20 m/s, at 120° to each of the others.

A 0.30 kg puck moving at 2.0 m/s [E] collides with a 0.20 kg puck and they stick, moving off at 1.6 m/s [E 30° N]. Find the initial velocity of the 0.20 kg puck.

Two objects of equal mass m and equal speed v, one moving east and one moving north, collide and stick. Show that exactly half of the kinetic energy is lost.

Check your understanding

Quiz

In a two-dimensional collision of an isolated system, conservation of momentum gives:

What is the size of the total momentum of 6.0 kg·m/s [E] and 8.0 kg·m/s [N]?

The direction [E 30° N] means:

An object at rest explodes into exactly two pieces. The pieces must move:

A puck bounces off smooth boards at the same speed and angle. Its change in momentum is:

What are the components of 20 kg·m/s [N 30° E] (x east, y north)?

For parents and teachers

Big idea (C2.5, C2.6): momentum is a vector. In two dimensions, conservation of momentum gives two equations, one for each component. Explosions from rest produce momenta that add to zero (a closed vector polygon).

How to help: The momentum map activity needs only paper, a ruler and a protractor. Check that your teen draws arrows tip to tail, uses a consistent scale, and labels directions.

Common difficulties:

Adding sizes instead of vectors (6.0 + 6.0 = 12 is wrong for perpendicular momenta).
Direction notation. [E 30° N] and [N 30° E] are different directions.
Signs of components. West and south components are negative; a quick sketch prevents most errors.
Calculator in degree mode, and choosing the quadrant from the signs of the components rather than trusting tan⁻¹ alone.