Further Mechanics — Momentum and collisions: A-level Further Maths OCR revision notes

6.03

Momentum and impulse

Momentum and impulse 6.03

Definitions

Momentum: $p = m v$ , a vector, measured in N s (equivalently kg m s⁻¹).
Impulse of a constant force: $I = F t$ . Of a variable force: $I = \int F d t$ .

Key results

Impulse–momentum principle: $I = m v - m u$ .
Conservation of linear momentum: if no external impulse acts on a system, its total momentum is unchanged.
For a variable force in one dimension, the impulse is the area under the force–time graph.

Method

Choose a positive direction and keep to it. Velocities in the opposite direction are negative.
In two dimensions, work with $i$ , $j$ components; the impulse–momentum principle holds for each component separately.

In practiceA ball of mass 0.2 kg moving with velocity $(4 i + 3 j)$ m s $^{- 1}$ receives an impulse of $(- 2 i + 0.4 j)$ N s. Find its new velocity and speed.

$I = m v - m u$ impulse equals the change in momentum
$0.2 v = 0.2 (4 i + 3 j) + (- 2 i + 0.4 j) = - 1.2 i + 1.0 j$ each component separately
$v = - 6 i + 5 j m s^{- 1}, ∣ v ∣ = 61 \approx 7.81 m s^{- 1}$

Notes

Impulse is a vector. An impulse of 4 N s in the wrong direction gives a different final velocity, so draw a diagram with arrows.
During an impact, weight and friction give negligible impulse compared with the contact force: this is why momentum is treated as conserved in collisions.

Worked example

A ball of mass 0.2 kg moving with velocity $(5 i + 2 j)$ m s⁻¹ receives an impulse of $(- 3 i + j)$ N s.

Find its new velocity and speed.

Show worked solution

0.2 v = 0.2 (5 i + 2 j) + (- 3 i + j) = - 2 i + 1.4 j

so:

v = (- 10 i + 7 j)

s⁻¹.

Speed:

= 149 = 12.2

s⁻¹.

6.03

Direct impacts and Newton's experimental law

Newton's experimental law and direct impacts 6.03

Definitions

Direct impact: a collision in which both velocities are along the line of centres.
Coefficient of restitution $e$ : $speed of separation = e \times speed of approach$ , with $0 \leq e \leq 1$ .

Key results

For spheres A and B, with B ahead in the positive direction: $v_{B} - v_{A} = e (u_{A} - u_{B})$ .
$e = 1$ : perfectly elastic, no kinetic energy lost. $e = 0$ : perfectly inelastic, the bodies coalesce.
Kinetic energy is lost in every collision with $e < 1$ , even though momentum is conserved.

Method

Draw before and after diagrams with all velocities in the positive direction, labelled as unknowns.
Write the momentum equation and the restitution equation, and solve them simultaneously.
Interpret signs: a negative velocity means motion in the negative direction.

In practiceA (2 kg, 5 m s $^{- 1}$ ) collides directly with B (3 kg, at rest), with $e = 0.4$ . Find the velocities after impact and the kinetic energy lost.

$2 (5) + 3 (0) = 2 v_{A} + 3 v_{B}$ momentum, both after-velocities taken positive
$v_{B} - v_{A} = 0.4 (5 - 0) = 2$ separation speed $= e \times$ approach speed
$10 = 2 v_{A} + 3 (v_{A} + 2) ⟹ v_{A} = 0.8, v_{B} = 2.8 m s^{- 1}$ both positive: both move on
$Δ E_{k} = \frac{1}{2} (2) (5^{2}) - [\frac{1}{2} (2) (0. 8^{2}) + \frac{1}{2} (3) (2. 8^{2})] = 25 - 12.4 = 12.6 J$

Notes

If the before and after directions are written correctly, the restitution equation's signs take care of themselves. Do not 'adjust' signs by intuition.
To decide whether further collisions occur, compare the velocities after each impact: a later collision needs the body behind to be moving faster in the same direction, or the two to be moving towards each other.

Impacts with a fixed surface 6.03

Key results

Rebound speed $= e \times$ impact speed, perpendicular to the surface.
A ball dropped from height $h_{0}$ rebounds to $e^{2} h_{0}$ , then $e^{4} h_{0}$ , …
Successive flight times form a geometric series with ratio $e$ ; their sum gives the total time before the ball comes to rest.

Method

Use $v^{2} = u^{2} + 2 a s$ for the impact speed, multiply by $e$ for the rebound, and repeat.
Sum the infinite geometric series of flight times: $t_{0} + \frac{2 e t _{0}}{1 - e}$ for a ball dropped from rest.

In practiceA ball is dropped from 5 m onto a floor with $e = 0.6$ . Find the first rebound height and the total time until it stops bouncing ( $g = 9.8$ ).

$u = 2 (9.8) (5) = 9.90 m s^{- 1}, v = e u = 5.94 m s^{- 1}$ impact, then rebound
$h_{1} = \frac{v ^{2}}{2 g} = e^{2} h_{0} = 0.36 \times 5 = 1.8 m$ each rebound height is $e^{2}$ times the last
$t_{0} = \frac{2 ( 5 )}{9.8} = 1.01 s$ the first fall
$T = t_{0} + \frac{2 e t _{0}}{1 - e} = 1.01 + \frac{2 ( 0.6 ) ( 1.01 )}{0.4} = 4.04 s$ each up-and-down takes $e$ times as long as the one before

Notes

The model predicts infinitely many bounces in a finite time, which is a limitation of the model, not of the arithmetic.
The impulse from the floor is $m (v + u)$ for impact speed $u$ and rebound speed $v$ , because the velocity reverses.

Worked examples

Worked example

Sphere A (1 kg) moving at 6 m s⁻¹ hits sphere B (2 kg) at rest, with:

e = \frac{1}{2}

B then hits a wall perpendicular to its motion, with:

e = \frac{1}{3}

Show that A and B collide again, and find their velocities afterwards.

Will there be a third collision?

Show worked solution

First impact: $6 = v_{A} + 2 v_{B}$ and:

v_{B} - v_{A} = \frac{1}{2} (6) = 3

so $v_{B} = 3$ , $v_{A} = 0$ .

B rebounds from the wall at:

\frac{1}{3} (3) = 1

s⁻¹ towards A, which is at rest, so they collide again.

Second impact (towards the wall positive):

2 (- 1) = v_{A} + 2 v_{B}

and:

v_{B} - v_{A} = \frac{1}{2} (0 - (- 1)) = \frac{1}{2}

Then:

3 v_{B} = - \frac{3}{2}

so:

v_{B} = - \frac{1}{2}

and $v_{A} = - 1$ .

Both now move away from the wall, A faster than B, so they separate: no third collision.

Worked example

A ball is dropped from 2 m onto a floor with $e = 0.6$ .

Find the height of the first rebound and the total time before it comes to rest. ( $g = 9.8$ m s⁻²)

Show worked solution

Impact speed:

2 (9.8) (2) = 6.26

s⁻¹, rebound speed $3.76$ m s⁻¹, so the rebound height is:

0. 6^{2} \times 2 = 0.72

The first fall takes:

t_{0} = 4/9.8 = 0.639

each later flight is $e$ times as long as the one before.

Total:

= 0.639 + \frac{2 ( 0.6 ) ( 0.639 )}{1 - 0.6} = 2.56

6.03

Oblique impacts

Oblique impact with a smooth surface 6.03

Key results

A smooth surface exerts an impulse only along the normal, so the velocity component parallel to the surface is unchanged.
The component perpendicular to the surface reverses and is multiplied by $e$ .
With angles measured from the surface: $tan β = e tan α$ .

Method

Resolve the incoming velocity parallel and perpendicular to the surface.
Apply 'parallel unchanged' and 'perpendicular × e', then recombine to find speed and direction.

In practiceA ball hits a smooth wall at 8 m s $^{- 1}$ , at $3 0^{\circ}$ to the wall, with $e = 0.5$ . Find its speed and direction afterwards.

$∥: 8 cos 3 0^{\circ} = 43, ⊥: 8 sin 3 0^{\circ} = 4$ resolve along and perpendicular to the wall
$∥ after = 43, ⊥ after = 0.5 \times 4 = 2$ parallel unchanged, perpendicular $\times e$
$speed = 48 + 4 = 52 = 7.21 m s^{- 1}$
$tan β = \frac{2}{4 3} ⟹ β = 16. 1^{\circ} to the wall$ check: $tan β = e tan 3 0^{\circ}$

Notes

Check whether angles are given from the surface or from the normal; $tan β = e tan α$ depends on which.
Since $e < 1$ , the rebound is closer to the surface and slower than the approach.

Oblique collisions of two smooth spheres 6.03

Definitions

Line of centres: the line through the centres of two spheres at the instant of impact. For smooth spheres, the impulse between them acts along it.

Key results

Perpendicular to the line of centres, each sphere's velocity component is unchanged.
Along the line of centres, momentum is conserved and Newton's law applies, exactly as in a direct impact.
Along the line of centres, with components $u_{1}, u_{2}$ before and $v_{1}, v_{2}$ after: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ and $v_{2} - v_{1} = e (u_{1} - u_{2})$ .
Example: sphere $A$ moving at $4 m s^{- 1}$ at $6 0^{\circ}$ to the line of centres hits an identical sphere $B$ at rest, with $e = \frac{1}{2}$ . Along the line, $u_{1} = 4 cos 6 0^{\circ} = 2$ , so $v_{1} + v_{2} = 2$ and $v_{2} - v_{1} = 1$ : $v_{1} = 0.5$ , $v_{2} = 1.5$ . $A$ keeps its perpendicular component $4 sin 6 0^{\circ} = 23$ , so it moves off at $0.25 + 12 = 3.5 m s^{- 1}$ ; $B$ moves along the line of centres at $1.5 m s^{- 1}$ .
Kinetic energy is lost unless $e = 1$ : in this example it falls from $8 m$ to $\frac{1}{2} m (12.25) + \frac{1}{2} m (2.25) = 7.25 m$ .

Method

Resolve each velocity along and perpendicular to the line of centres.
Solve the direct-impact problem along the line of centres; carry the perpendicular components across unchanged.
Recombine components to give speeds and directions.

In practiceEqual smooth spheres A and B have mass 1 kg. A moves at 6 m s $^{- 1}$ at $6 0^{\circ}$ to the line of centres and hits B, at rest; $e = \frac{1}{2}$ . Find the velocities after impact.

$A : 6 cos 6 0^{\circ} = 3 along, 6 sin 6 0^{\circ} = 33 perpendicular$ resolve
$3 = v_{A} + v_{B}, v_{B} - v_{A} = \frac{1}{2} (3)$ momentum and restitution along the line of centres
$v_{A} = 0.75, v_{B} = 2.25 m s^{- 1}$
$A : 0.7 5^{2} + (33)^{2} = 5.25 m s^{- 1} at arctan \frac{3 3}{0.75} = 81. 8^{\circ} to the line of centres$ its perpendicular part is unchanged
B moves off along the line of centres at 2.25 m s $^{- 1}$ , as a sphere at rest always does.

Notes

A sphere at rest before impact moves off along the line of centres, because it receives no perpendicular impulse.
For equal masses and $e = 1$ , the two spheres separate at right angles when one was at rest.

Worked examples

Worked example

A ball moving at 10 m s⁻¹ hits a smooth wall at 50° to the wall, with:

e = \frac{1}{2}

Find its speed and direction after impact, and the fraction of kinetic energy lost.

Show worked solution

Parallel:

10 cos 5 0^{\circ} = 6.43

s⁻¹, unchanged.

Perpendicular:

\frac{1}{2} (10 sin 5 0^{\circ}) = 3.83

s⁻¹.

Speed:

= 6.4 3^{2} + 3.8 3^{2} = 7.48

s⁻¹, at:

tan^{- 1} (3.83/6.43) = 30. 8^{\circ}

to the wall.

KE fraction lost:

= 1 - \frac{7.4 8 ^{2}}{100} = 0.44

Worked example

Two equal smooth spheres; A moves at 4 m s⁻¹ at 60° to the line of centres and strikes B, which is at rest, with:

e = \frac{1}{2}

Find the velocities after impact.

Show worked solution

Perpendicular: A keeps $4 sin 6 0^{\circ} = 23$ ; B has none.

Along the line of centres, A's component is $4 cos 6 0^{\circ} = 2$ : $2 = v_{A} + v_{B}$ and:

v_{B} - v_{A} = \frac{1}{2} (2) = 1

so $v_{A} = 0.5$ , $v_{B} = 1.5$ .

B moves at 1.5 m s⁻¹ along the line of centres.

A moves at:

0.25 + 12 = 3.5

s⁻¹, at:

tan^{- 1} (23 /0.5) = 81. 8^{\circ}

to the line of centres.