Prerequisites
Before studying this topic, review:
- Work Done by Forces - Definition of work
What is Kinetic Energy?
Kinetic energy is the energy possessed by a body due to its motion. It’s a scalar quantity that depends on both mass and speed.
Definition
$$\boxed{KE = \frac{1}{2}mv^2}$$where:
- $m$ = mass of the body (kg)
- $v$ = speed of the body (m/s)
- KE = kinetic energy (Joules)
- Always positive (since $v^2 \geq 0$)
- Scalar quantity (no direction)
- Zero when body is at rest
- Depends on reference frame
Interactive Demo: Visualize Energy Changes
See how kinetic energy changes with velocity and mass.
Derivation
Starting from the work-energy perspective:
Consider a body of mass $m$ accelerated from rest by a constant force $F$.
- From Newton’s second law: $F = ma$
- Using $v^2 = u^2 + 2as$ (from kinematic equations) with $u = 0$: $v^2 = 2as$, so $s = \frac{v^2}{2a}$
- Work done: $W = Fs = ma \cdot \frac{v^2}{2a} = \frac{1}{2}mv^2$
This work done equals the kinetic energy acquired.
Kinetic Energy and Momentum
There’s an important relationship between KE and momentum $p = mv$:
$$\boxed{KE = \frac{p^2}{2m}}$$Proof
$$KE = \frac{1}{2}mv^2 = \frac{1}{2m}(mv)^2 = \frac{p^2}{2m}$$Applications
This relation is useful when:
- Momentum is conserved but velocity isn’t known
- Comparing energies of bodies with same momentum
For same momentum:
$$KE \propto \frac{1}{m}$$Lighter body has more kinetic energy
For same kinetic energy:
$$p \propto \sqrt{m}$$Heavier body has more momentum
Change in Kinetic Energy
When speed changes from $v_1$ to $v_2$:
$$\Delta KE = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \frac{1}{2}m(v_2^2 - v_1^2)$$Using difference of squares:
$$\Delta KE = \frac{1}{2}m(v_2 + v_1)(v_2 - v_1)$$Percentage Change in KE
If speed changes by factor $(1 + x)$ where $x$ is fractional change:
$$\frac{\Delta KE}{KE} = (1+x)^2 - 1 = 2x + x^2$$For small $x$:
$$\frac{\Delta KE}{KE} \approx 2x$$If speed doubles ($x = 1$), KE increases by factor of 4, not 2!
$$KE' = \frac{1}{2}m(2v)^2 = 4 \times \frac{1}{2}mv^2 = 4KE$$Kinetic Energy in Different Scenarios
1. Rotational KE
For a rotating body with moment of inertia $I$ and angular velocity $\omega$:
$$KE_{rot} = \frac{1}{2}I\omega^2$$2. Rolling Motion
For a rolling body (both translation + rotation):
$$KE_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$3. Relativistic KE (For Reference)
At speeds close to light:
$$KE = (\gamma - 1)mc^2$$where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$
Worked Examples
Example 1: Comparing Kinetic Energies
Solution: For same momentum $p$:
$$\frac{KE_{car}}{KE_{motorcycle}} = \frac{p^2/2m_{car}}{p^2/2m_{motorcycle}} = \frac{m_{motorcycle}}{m_{car}} = \frac{200}{1000} = \boxed{1:5}$$The motorcycle has 5 times more kinetic energy!
Example 2: Speed Change
Solution: Let initial speed = $v$, final speed = $1.2v$
$$\frac{\Delta KE}{KE} = \frac{(1.2v)^2 - v^2}{v^2} = \frac{1.44 - 1}{1} = 0.44 = \boxed{44\%}$$Summary
| Quantity | Formula | Depends On |
|---|---|---|
| Kinetic Energy | $\frac{1}{2}mv^2$ | Mass, Speed |
| KE from momentum | $\frac{p^2}{2m}$ | Momentum, Mass |
| Change in KE | $\frac{1}{2}m(v_f^2 - v_i^2)$ | Initial & final speeds |
Practice Problems
A bullet (20 g) moving at 500 m/s and a truck (5000 kg) moving at 1 m/s. Compare their kinetic energies and momenta.
Two bodies A (3 kg) and B (4 kg) have the same kinetic energy. Find the ratio of their momenta.
A body’s momentum increases by 50%. What’s the percentage change in kinetic energy?