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C. Energy and Momentum · Momentum and Collisions

Glancing Collisions in Two Dimensions

C2.6 C3.3 C2.5About 70 minutes
Learning goals
  • I can solve two-dimensional collisions using conservation of momentum in x and y.
  • I can decide whether a two-dimensional collision is elastic or inelastic.
  • I can explain and use the 90° rule for elastic glancing collisions between equal masses.
  • I can reconstruct the speeds of vehicles in an intersection collision.

Words to know

glancing (oblique) collision
A collision in which the objects strike off-centre and move off in different directions.
line of centres
The line joining the centres of two round objects at the moment they touch; the contact force acts along it.
90° rule
For equal masses, one initially at rest, an elastic collision sends the objects off at right angles to each other.

Off-centre hits

In curling, a shooter stone that hits a stationary stone off-centre sends it off at an angle, and the shooter itself veers away in another direction. Billiards, air hockey and collisions between atoms work the same way. A collision in which the objects do not move along a single line is called a glancing (or oblique) collision.

In the last lesson you used conservation of momentum in two dimensions for explosions and sticking collisions. Now you will analyse glancing collisions, decide whether they are elastic, and discover a surprising rule for equal masses.

Analysing a glancing collision

For two smooth, round objects, the collision force acts along the line joining their centres at the moment of contact. So a target at rest moves off along the line of centres. How far off-centre the hit is decides the target's direction.

To solve a glancing collision:

  1. Choose axes. Usually +x is the direction of the incoming object, so all the initial momentum is in x and the initial y-momentum is zero.
  2. Write two momentum equations, one for x and one for y.
  3. If the collision is elastic, you also have a kinetic energy equation (a third equation).
  4. Solve, then check by recomputing kinetic energy to classify the collision.

A special rule: equal masses, elastic, target at rest

For equal masses with the target at rest, the two objects always move off at 90° to each other after an elastic glancing collision.

Why: momentum gives the vector equation mv = mv₁′ + mv₂′, so v = v₁′ + v₂′: the three velocity vectors form a triangle. Kinetic energy gives ½mv² = ½mv₁′² + ½mv₂′², so v² = v₁′² + v₂′². That is the Pythagorean theorem, which is true only for a right triangle. So the angle between v₁′ and v₂′ is 90°. (Using the dot product from MCV4U: v² = v₁′² + v₂′² + 2v₁′·v₂′, so v₁′·v₂′ = 0.)

Billiards players use this rule constantly: after a clean hit with no spin, the cue ball leaves at right angles to the object ball. Curling stones are close to elastic and of equal mass, so the rule works roughly for them too (their rotation and the ice change it a little).

shooter arrives: 2.0 m/sline of centrestarget 1.73 m/sshooter 1.0 m/s30° above and60° below theoriginal line:90° apart
An off-centre hit between two equal curling stones (seen from above). The target moves along the line of centres, here 30° above the shooter's original path. If the collision is elastic, the shooter moves off 60° below the original line, so the two paths are 90° apart.
Worked example

Example 1: Curling stones

A curling stone moving at 2.0 m/s hits an identical stone at rest. The target stone moves off at 30° to the original direction. Assuming an elastic collision, find both final velocities.

  1. Elastic, equal masses: the paths are 90° apart, so the shooter moves off at 60° on the other side of the original line.
  2. Let +x be the original direction, and let the target go above the x-axis.
  3. y-momentum: 0 = v₂′ sin 30° − v₁′ sin 60°, so v₁′ = v₂′ sin 30° ÷ sin 60°.
  4. x-momentum: 2.0 = v₂′ cos 30° + v₁′ cos 60°. Substituting: 2.0 = v₂′(cos 30° + sin 30° cos 60° ÷ sin 60°) = v₂′(0.866 + 0.289) = 1.155v₂′, so v₂′ ≈ 1.73 m/s.
  5. v₁′ = 1.73 × 0.5 ÷ 0.866 ≈ 1.0 m/s. (Shortcut: the right-triangle in the next figure gives v₂′ = 2.0 cos 30° and v₁′ = 2.0 cos 60°.)
  6. Check kinetic energy: 2.0² = 4.0, and 1.73² + 1.0² = 3.0 + 1.0 = 4.0 ✓.

Target: 1.73 m/s at 30° to one side. Shooter: 1.0 m/s at 60° to the other side.

p before: m(2.0) along original linetarget: m(1.73)shooter: m(1.0)Elastic, equal masses:v² = v₁′² + v₂′²(Pythagoras), so thetriangle has a right anglebetween the two final paths.
The momentum triangle for Example 1. The initial momentum equals the sum of the two final momenta. For equal masses in an elastic collision, the triangle has a right angle between the two final momenta.
Worked example

Example 2: Unequal pucks

A 0.20 kg puck moving at 4.0 m/s [E] hits a 0.30 kg puck at rest. Afterwards the 0.20 kg puck moves at 2.0 m/s [E 50° N]. Find the velocity of the 0.30 kg puck and classify the collision.

  1. x: (0.20)(4.0) = (0.20)(2.0)cos 50° + 0.30v_x, so 0.80 = 0.257 + 0.30v_x and v_x ≈ 1.81 m/s.
  2. y: 0 = (0.20)(2.0)sin 50° + 0.30v_y, so 0.30v_y = −0.306 and v_y ≈ −1.02 m/s.
  3. Speed: √(1.81² + 1.02²) ≈ 2.08 m/s. Direction: tan⁻¹(1.02 ÷ 1.81) ≈ 29.4° south of east.
  4. Ek before = ½(0.20)(4.0)² = 1.60 J. Ek after = ½(0.20)(2.0)² + ½(0.30)(2.08)² ≈ 0.40 + 0.65 = 1.05 J.

The 0.30 kg puck moves at about 2.1 m/s [E 29.4° S]. Kinetic energy fell from 1.60 J to about 1.05 J (about 34% lost), so the collision is inelastic.

Worked example

Example 3: Two unknown speeds, known directions

A 0.50 kg puck moving at 6.0 m/s [E] hits an identical puck at rest. Photos show that afterward the first puck moves [E 40° N] and the second moves [E 50° S]. Find both speeds and classify the collision.

  1. y: 0 = 0.50v₁′ sin 40° − 0.50v₂′ sin 50°, so v₂′ = v₁′ sin 40° ÷ sin 50°.
  2. x: 6.0 = v₁′ cos 40° + v₂′ cos 50° (the masses cancel). Substituting: 6.0 = v₁′(cos 40° + sin 40° cos 50° ÷ sin 50°) = v₁′(0.766 + 0.539) = 1.305v₁′.
  3. v₁′ ≈ 4.60 m/s and v₂′ ≈ 4.60 × 0.643 ÷ 0.766 ≈ 3.86 m/s.
  4. Kinetic energy (per ½m): before 36; after 4.60² + 3.86² ≈ 21.1 + 14.9 = 36.0 ✓.

4.6 m/s [E 40° N] and 3.9 m/s [E 50° S]. Kinetic energy is conserved: the collision is elastic, as the 90° angle between the paths suggested.

Worked example

Example 4: Reconstructing an intersection crash

A 1200 kg car travelling east and a 1600 kg truck travelling north collide in an intersection and lock together. Skid analysis shows the wreck moved off at 12 m/s [E 40° N]. Find each vehicle's speed before the crash.

  1. Total momentum after: (2800)(12) = 33 600 kg·m/s at 40° north of east.
  2. x (only the car had east momentum): 1200v_car = 33 600 cos 40° ≈ 25 739, so v_car ≈ 21.4 m/s.
  3. y (only the truck had north momentum): 1600v_truck = 33 600 sin 40° ≈ 21 598, so v_truck ≈ 13.5 m/s.
  4. In km/h: 21.4 × 3.6 ≈ 77 km/h and 13.5 × 3.6 ≈ 49 km/h.

The car was moving about 21 m/s (77 km/h) and the truck about 13.5 m/s (49 km/h). The direction of the wreck tells you the ratio of the two momenta; its speed tells you their size.

Just for fun

Many championship curling stones are made from granite quarried on Ailsa Craig, a tiny, uninhabited island off the coast of Scotland. So when Canadians curl, there is a good chance two pieces of a Scottish island are having a glancing collision.

Hands-on activity

Glancing coins

You need: two identical coins (loonies work well), a large sheet of paper taped to a smooth table, a pencil, a ruler and a protractor.

  1. Place the target coin on the paper and trace around it.
  2. Place the shooter coin about 15 cm away and trace it. Flick the shooter so it hits the target off-centre.
  3. Trace both coins where they stop. Draw straight lines from each coin's position at the collision (the target's start, and the point where the shooter touched it) to where it stopped.
  4. Measure the angle between the two paths after the collision. Repeat for several different off-centre hits.
  5. Bonus: friction stops each coin over a distance that depends on its speed squared, so the ratio of the distances slid gives the ratio of the speeds squared. Use this to check whether kinetic energy was roughly conserved.

What to look for: the angle between the paths is usually somewhat less than 90° (often 70° to 85°). Coins are not perfectly elastic, they spin, and friction acts during the hit, so the 90° rule is only approximate. A head-on hit gives the shooter stopping and the target going straight on.

Your turn

1

A cue ball moving at 3.0 m/s hits an identical ball at rest. The object ball leaves at 40° to the cue ball's original path. Assuming an elastic collision, find the speeds and directions of both balls.

2

Two identical pucks have an elastic glancing collision, with one initially at rest. What is the angle between their paths afterward?

3

A 0.20 kg puck moving at 4.0 m/s [E] hits an identical puck at rest. Afterward the first puck moves at 2.0 m/s [E 60° N]. Find the second puck's velocity and classify the collision.

4

A 1000 kg car travelling east and a 1500 kg van travelling north lock together and move off at 10 m/s [E 45° N]. Find their speeds before the crash.

5

Find the percentage of kinetic energy lost in the previous question.

6

A student claims two identical billiard balls left an elastic collision at 30° on each side of the original line. Explain why this is impossible.

More practice: warm-up

1

What is a glancing collision?

2

In which direction does a smooth target ball at rest move after being hit?

3

For equal masses in an elastic glancing collision with the target at rest, what is the angle between the final paths?

4

How many equations does conservation of momentum give in a two-dimensional collision?

5

If the collision is also known to be elastic, how many equations do you have?

More practice: on your own

1

A 2.0 kg ball moving at 5.0 m/s [E] hits a 3.0 kg ball at rest. Afterward the 2.0 kg ball moves at 2.0 m/s [N]. Find the 3.0 kg ball's velocity and classify the collision.

2

A curling stone moving at 1.5 m/s [E] hits an identical stone at rest. The target moves at 1.2 m/s [E 36.9° N]. Find the shooter's velocity and decide whether the collision was elastic.

3

Give two reasons why real billiard or curling collisions do not give exactly 90° between the paths.

4

A puck moving at 3.0 m/s [N] hits an identical puck at rest. Afterward the first puck moves at 1.5 m/s [N 60° W]. Find the second puck's velocity.

5

A 0.10 kg puck moving at 6.0 m/s [E] hits a puck of unknown mass at rest. The 0.10 kg puck moves off at 3.0 m/s [N], and the other puck moves at 2.0 m/s. Find the unknown mass and the second puck's direction.

Challenge

1

Using vectors, prove that for two equal masses in an elastic collision with one at rest, the final velocities are perpendicular (unless one is zero).

2

A 1500 kg car travelling north and a 1000 kg car travelling east collide and lock together. The wreck skids 15 m in the direction [N 30° E] on a road with μk = 0.60. Find each car's speed before the collision.

3

Two identical pucks: one moving at 4.0 m/s hits the other at rest. Afterward both move at 2.5 m/s at equal angles on either side of the original line. Find the angle and the percentage of kinetic energy lost.

Check your understanding

Quiz

1.

Two identical pucks have an elastic glancing collision. One was at rest. Afterward, the angle between their paths is:

2.

A puck moving east hits a puck at rest, off-centre. After the collision, the total north-south momentum of the two pucks is:

3.

After a glancing collision, you know the masses, the initial velocities and the final velocity of one object. Which conservation law lets you find the other object's final velocity?

4.

Two identical billiard balls, one at rest, move off at 70° to each other after a collision. This suggests the collision was:

5.

In an intersection crash where the vehicles lock together, the direction in which the wreck moves tells investigators:

6.

A smooth ball at rest is struck off-centre. It moves off:

For parents and teachers

Big idea (C2.6, C3.3): in a glancing collision, momentum is conserved separately in x and y. If the collision is also elastic, kinetic energy gives a third equation. For equal masses with one at rest, an elastic glancing collision always sends the two objects off at 90° to each other; a smaller angle means kinetic energy was lost.

How to help: Try the glancing-coins activity together and measure the angle. Ask your teen why it is close to, but usually below, 90°.

Common difficulties:

  • Setting up axes. Putting +x along the incoming direction makes the initial y-momentum zero, which simplifies everything.
  • Sign of the y-components. Objects that go to opposite sides of the line must have opposite signs of y-momentum.
  • Assuming elastic without checking. Always compute the kinetic energy before and after.
  • Calculator mode. Degrees, not radians.