Kinetic Energy: notes and previous year questions

1The energy of motion

Kinetic energy is the energy a body has because it moves. It is the work needed to bring the body from rest up to its speed, and also the work it can do on other things while it is being stopped.

• It is never negative, because $v^2$ cannot be negative.
• It is a scalar: it has no direction.
• It is zero when the body is at rest.
• It depends on the frame: a passenger has no kinetic energy relative to the train, but plenty relative to the platform.

2Where ½mv² comes from

Push a body of mass $m$ from rest with a steady force $F$.

1. Second law: $F = ma$.
2. From rest, $v^2 = 2as$, so $s = v^2/2a$.
3. The work done is $W = Fs = ma \cdot \dfrac{v^2}{2a} = \tfrac{1}{2}mv^2$.

With calculus it works for any force. Using the chain rule $\frac{dv}{dt} = v\frac{dv}{ds}$:

3Speed enters squared

Because the speed is squared, doubling the speed makes the kinetic energy four times as big. Two cars braking with the same force: the one going twice as fast skids four times as far.

4Kinetic energy and momentum

Multiply the top and bottom of $\tfrac{1}{2}mv^2$ by $m$: $\tfrac{1}{2}mv^2 = \frac{(mv)^2}{2m}$. With $p = mv$:

• Same momentum: $KE \propto 1/m$, so the lighter body has more energy. A 1000 kg car at 10 m/s and a 200 kg motorcycle at 50 m/s both have 10 000 kg m/s, but the motorcycle has 5 times the kinetic energy (250 kJ against 50 kJ).
• Same kinetic energy: $p \propto \sqrt m$, so the heavier body has more momentum. 3 kg and 4 kg bodies with equal energy have momenta in the ratio $\sqrt 3 : 2$.

Because $KE \propto p^2$, a change in momentum is squared too: 50% more momentum means $1.5^2 = 2.25$ times the energy, an increase of 125%.

In short: momentum tells you how hard a body is to stop; kinetic energy tells you how much work it can do while it stops.

5A scalar that is never negative

Two identical 1000 kg cars drive toward each other at 10 m/s. Their momenta, +10 000 and −10 000 kg m/s, cancel. Their kinetic energies, 50 000 J each, add to 100 000 J. Kinetic energies never cancel.

A 60 kg passenger in a train at 20 m/s has no kinetic energy relative to the train, and $\tfrac{1}{2}(60)(20)^2 = 12\,000$ J relative to the platform.

6Kinetic energy against time and distance

A steady force $F$ acts on a mass $m$ from rest. Then $v = Ft/m$, so:

Example: 10 N on 2 kg from rest gives $KE = 25t^2$, so it reaches 100 J at $t = 2$ s. If the energy is 20 J after 2 s, it is 80 J after 4 s.

7Spinning and rolling

A spinning body has rotational kinetic energy. The moment of inertia $I$ plays the part of the mass, and the angular speed $\omega$ plays the part of the speed.

A rolling ball both moves along and spins, so it has both kinds:

At speeds close to light, Einstein's formula $KE = (\gamma - 1)mc^2$ with $\gamma = 1/\sqrt{1 - v^2/c^2}$ takes over. For everyday speeds it becomes exactly $\tfrac{1}{2}mv^2$.

8A puzzle: the man and the boy

A running man has half the kinetic energy of a boy with half the man's mass. When the man speeds up by 1 m/s, the two kinetic energies become equal. Find their speeds.

1. Man: mass $m$, speed $v_m$. Boy: mass $m/2$, speed $v_b$. The boy's energy is $\tfrac{1}{2}\cdot\tfrac{m}{2}v_b^2 = \tfrac{1}{4}mv_b^2$.
2. The man has half of that: $\tfrac{1}{2}mv_m^2 = \tfrac{1}{8}mv_b^2$, so $v_b = 2v_m$.
3. After speeding up: $\tfrac{1}{2}m(v_m + 1)^2 = \tfrac{1}{4}mv_b^2 = mv_m^2$, so $(v_m + 1)^2 = 2v_m^2$.
4. $v_m + 1 = \sqrt 2\,v_m$, so $v_m = \sqrt 2 + 1 \approx 2.41$ m/s and $v_b \approx 4.83$ m/s.

Check: the man has $\tfrac{1}{2}m(2.41)^2 \approx 2.9m$ and the boy $\tfrac{1}{4}m(4.83)^2 \approx 5.8m$, about twice as much.