Conservation of Linear Momentum and Collision

Lessons

Senior 4 · Physics

Conservation of Linear Momentum and Collision

4 lessons

Lesson 2 of 4

Kinetic and Potential Energy and the Conservation of Mechanical Energy

25 min

Why doubling your speed is so dangerous

A driver who doubles her speed from 50 km/h to 100 km/h does not need twice the distance to stop: on the same road with the same brakes she needs four times as far. A pendulum bob released from one height swings up to almost the same height on the other side, never higher. Both facts follow from two ideas you will derive in this lesson: the work–energy theorem and the conservation of mechanical energy.

Kinetic energy and the work–energy theorem

Let a constant resultant force $F$ act on a body of mass $m$ along its direction of motion while it moves a distance $s$ , speeding up from $u$ to $v$ . The net work done is $W = F s = m a s$ . From the equation of uniformly accelerated motion $v^{2} = u^{2} + 2 a s$ , we have $as = \frac{v^2 - u^2}{2}$ , so

W_{\text{net}} = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = \Delta E_k, \qquad E_k = \tfrac{1}{2}mv^2

This is the work–energy theorem: the net work done on a body equals the change in its kinetic energy. It holds for any force, constant or not, because a varying force can be split into many small steps in each of which it is constant. Negative net work (from brakes or friction) reduces the kinetic energy.

Definition

The kinetic energy of a body of mass $m$ moving at speed $v$ is $E_k = \tfrac{1}{2}mv^2$ . The work–energy theorem states that the net work done on a body equals its change in kinetic energy: $W_{\text{net}} = \Delta E_k$ . Example: a 0.40 kg football kicked at $20\,\mathrm{m\,s^{-1}}$ has $E_k = \tfrac{1}{2} \times 0.40 \times 20^2 = 80\,\mathrm{J}$ .

Worked example 1: stopping distance

A 1000 kg car travelling at $20\,\mathrm{m\,s^{-1}}$ brakes with a constant braking force of 8000 N. Find its braking distance, and the distance at $40\,\mathrm{m\,s^{-1}}$ .

-Fs = 0 - \tfrac{1}{2}mv^2 \;\Rightarrow\; s = \frac{mv^2}{2F} = \frac{1000 \times 20^2}{2 \times 8000} = 25\,\mathrm{m}

\text{at } 40\,\mathrm{m\,s^{-1}}:\quad s = \frac{1000 \times 40^2}{2 \times 8000} = 100\,\mathrm{m}

Braking distance is proportional to $v^{2}$ : twice the speed, four times the distance. That is why speed limits near schools are so low.

Potential energy

To lift a body of mass $m$ at steady speed through a height $h$ near the Earth's surface, you must apply an upward force $m g$ and do work $m g h$ . That work is stored as gravitational potential energy:

\Delta E_p = mgh

Only changes in potential energy matter, so we may choose any convenient level as zero (the floor, the ground, the lowest point of a swing). The formula assumes $g$ is constant, which is true for heights small compared with the Earth's radius. A stretched spring stores elastic potential energy $\tfrac{1}{2}kx^2$ (lesson 1).

Gravity and the spring force are conservative forces: the work they do on a body moving between two points does not depend on the path taken, and is fully recovered when the body returns. Friction and air resistance are non-conservative: the work they do depends on the path, and it turns mechanical energy into internal (thermal) energy that cannot be recovered as work.

Conservation of mechanical energy

When only conservative forces do work, the work done by gravity on a falling body is $W = mg(h_1 - h_2) = -\Delta E_p$ . By the work–energy theorem this equals $\Delta E_k$ . So $\Delta E_k + \Delta E_p = 0$ : the total mechanical energy stays constant.

\tfrac{1}{2}mv_1^2 + mgh_1 = \tfrac{1}{2}mv_2^2 + mgh_2

Figure 2.1: A falling 1.0 kg ball: potential energy turns into kinetic energy, and the total stays the same.

When non-conservative forces such as friction also act, the work $W_{nc}$ they do changes the mechanical energy:

W_{nc} = \Delta E_k + \Delta E_p

The mechanical energy "lost" is not destroyed: it becomes internal energy of the surfaces and the air. The total energy of an isolated system is always conserved.

Definition

The principle of conservation of mechanical energy: if only conservative forces (such as gravity and elastic forces) do work on a system, the sum of its kinetic and potential energies remains constant. A conservative force is one whose work between two points does not depend on the path. Example: a stone thrown upwards at $12\,\mathrm{m\,s^{-1}}$ rises to $h = \frac{v^2}{2g} = 7.3\,\mathrm{m}$ whatever its mass.

Worked example 2: a swinging pendulum

A pendulum bob is pulled aside so that it is 0.45 m above its lowest point, and released. Ignoring air resistance, find its speed (a) at the lowest point, (b) when it is 0.20 m above the lowest point.

\text{(a)}\;\; mgh = \tfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.45} = 3.0\,\mathrm{m\,s^{-1}}

\text{(b)}\;\; v = \sqrt{2g(0.45 - 0.20)} = \sqrt{2 \times 9.8 \times 0.25} = 2.2\,\mathrm{m\,s^{-1}}

The mass cancels: every pendulum released from the same height has the same speed at the bottom.

Loop-the-loop

Figure 2.2: A ball runs down a smooth track into a vertical loop.

To stay on the track at the top of the loop, the ball needs a centripetal force of at least $\frac{mv^2}{r}$ (S3, Unit 1). The smallest possible speed at the top is when the weight alone provides it: $mg = \frac{mv^2}{r}$ , so $v^{2} = g r$ . Conservation of energy from the release point (height $h$ ) to the top (height $2 r$ ):

mgh = mg(2r) + \tfrac{1}{2}m(gr) \;\Rightarrow\; h = 2.5\,r

The ball must start at least two and a half times the loop's radius above the bottom. Real roller coasters start higher, to allow for friction, and use loops that are tighter at the top.

Worked example 3: friction on a slide

A 25 kg child slides from rest down a playground slide whose top is 3.0 m above its bottom. She reaches the bottom at $5.0\,\mathrm{m\,s^{-1}}$ . The slide is 6.0 m long. Find the energy turned into internal energy and the average friction force.

\text{loss} = mgh - \tfrac{1}{2}mv^2 = 25 \times 9.8 \times 3.0 - \tfrac{1}{2} \times 25 \times 5.0^2 = 735 - 312.5 = 4.2 \times 10^{2}\,\mathrm{J}

W_{nc} = -f s \;\Rightarrow\; f = \frac{422.5}{6.0} = 70\,\mathrm{N}

Without friction she would have reached $\sqrt{2 \times 9.8 \times 3.0} = 7.7\,\mathrm{m\,s^{-1}}$ .

Experiment 2.1: testing energy conservation with a pendulum

Apparatus: a heavy bob (diameter $d$ about 4 cm) on a string about 1.5 m long, a clamp stand, a metre rule, a phone that records slow-motion video (or a light gate and timer).

Pull the bob aside and measure its height $h$ above its lowest position.
Release it and record the bob passing a vertical marker at the lowest point. From the video, find the time $\Delta t$ the bob takes to pass the marker; its speed is $v = \frac{d}{\Delta t}$ .
Repeat for several heights and plot $v^{2}$ against $h$ . Energy conservation predicts a straight line through the origin with gradient $2 g$ .

$h$ (m)	0.10	0.20	0.30	0.40
$v$ ( $\mathrm{m\,s^{-1}}$ )	1.38	1.96	2.40	2.76
$v^{2}$ ( $\mathrm{m^2\,s^{-2}}$ )	1.90	3.84	5.76	7.62

Table 2.1: Illustrative results. The best line through the origin has gradient $19.1\,\mathrm{m\,s^{-2}}$ , compared with $2g = 19.6\,\mathrm{m\,s^{-2}}$ : about 97% of the potential energy appears as kinetic energy; air resistance and friction at the support account for the rest.

Warning

Safety. Keep heads clear of the swing of a heavy bob, and make sure the clamp stand cannot tip over (weigh down its base).

Warning

Common mistakes. (1) Using $m g h$ with $h$ as a distance along a slope: it is the vertical height. (2) Forgetting to square the speed in $\tfrac{1}{2}mv^2$ . (3) Applying conservation of mechanical energy when friction acts: use $W_{nc} = \Delta E_k + \Delta E_p$ . (4) Saying energy is "lost" in friction as if destroyed: it is transferred to internal energy.

Tip

Energy methods ignore the path and the time. If a question asks only for a speed at a height (not for a time or an acceleration), write "energy at A = energy at B" and solve; it is usually shorter than using equations of motion.

A classic physics demonstration with a pendulum (Rhett Allain). Watch a heavy pendulum released from rest at someone's chin: it never swings back higher than its starting point, because its mechanical energy cannot increase.

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Practice

Find the kinetic energy of a 1500 kg minibus moving at $15\,\mathrm{m\,s^{-1}}$ .
A 0.50 kg ball is dropped from 20 m. Find its speed when it is 5.0 m above the ground (ignore air resistance).
A stone is thrown vertically upwards at $12\,\mathrm{m\,s^{-1}}$ . How high does it rise?
A spring of stiffness $400\,\mathrm{N\,m^{-1}}$ is compressed 10 cm and used to launch a 20 g ball horizontally. Find the launch speed.
A 60 kg cyclist freewheels from rest down a hill 15 m high and reaches the bottom at $12\,\mathrm{m\,s^{-1}}$ . How much energy was transferred to internal energy by friction and air resistance?
A car's braking distance is 18 m at $15\,\mathrm{m\,s^{-1}}$ . Estimate its braking distance at $25\,\mathrm{m\,s^{-1}}$ with the same braking force.
A student says that a ball rolled down a frictionless slope reaches the bottom faster than one dropped straight down from the same height. Evaluate this statement for both speed and time.

Show answersHide answers

$E_k = \tfrac{1}{2} \times 1500 \times 15^2 = 1.7 \times 10^{5}\,\mathrm{J}$ .
It has fallen 15 m: $v = \sqrt{2 \times 9.8 \times 15} = 17\,\mathrm{m\,s^{-1}}$ .
$h = \frac{12^2}{2 \times 9.8} = 7.3\,\mathrm{m}$ .
$\tfrac{1}{2} \times 400 \times 0.10^2 = \tfrac{1}{2} \times 0.020 \times v^2$ , so $v = \sqrt{\frac{400 \times 0.010}{0.020}} = 14\,\mathrm{m\,s^{-1}}$ .
$mgh - \tfrac{1}{2}mv^2 = 60 \times 9.8 \times 15 - \tfrac{1}{2} \times 60 \times 12^2 = 8820 - 4320 = 4.5 \times 10^{3}\,\mathrm{J}$ .
$s \propto v^2$ : $18 \times \left(\frac{25}{15}\right)^2 = 50\,\mathrm{m}$ .
Speed: both arrive with the same speed $\sqrt{2gh}$ , because the same potential energy becomes kinetic energy and the normal force of the slope does no work. Time: the ball on the slope takes longer, because its acceleration $g\sin\theta$ is smaller and its path is longer. So the statement is wrong on both counts.

Quick check

A car's speed doubles. By what factor does its kinetic energy increase?

Key points

$E_k = \tfrac{1}{2}mv^2$ ; work–energy theorem $W_{\text{net}} = \Delta E_k$ ; stopping distance $\propto v^2$ .
$\Delta E_p = mgh$ (gravity), $\tfrac{1}{2}kx^2$ (spring); gravity and springs are conservative forces.
Only conservative forces: $E_k + E_p = \text{constant}$ . With friction: $W_{nc} = \Delta E_k + \Delta E_p$ .
Loop-the-loop needs $h \ge 2.5r$ .

Sources and credits

7.2: Kinetic Energy and the Work-Energy Theorem, 7.4: Conservative Forces and Potential Energy and 7.5: Nonconservative Forces, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Loop Coaster MOMOnGA vertical loop - Yomiuriland.jpg by Jeremy Thompson, Wikimedia Commons, licensed under CC BY 2.0.
Video: Classic Physics Demo With a Pendulum, Rhett Allain, embedded from YouTube.

The derivations, the two diagrams, the experiment and its illustrative results, the worked examples and the practice exercises are written by Ganzaa.

1Why doubling your speed is so dangerous

2Kinetic energy and the work–energy theorem

3Potential energy

4Conservation of mechanical energy

5Loop-the-loop

6Experiment 2.1: testing energy conservation with a pendulum