Power measures how fast work is done. Two cranes may lift the same load to the same height — doing identical work — but the one that does it in less time is more powerful. Power is defined as the rate of doing work, $P=\frac{W}{t}$. The SI unit is the watt (W), where $1\ \text{W}=1\ \text{J/s}$. A common practical unit is the horsepower, $1\ \text{hp}=746\ \text{W}$.
Instantaneous power can be written in terms of force and velocity. Since $dW=\vec{F}\cdot d\vec{s}$, dividing by $dt$ gives $P=\vec{F}\cdot\frac{d\vec{s}}{dt}=\vec{F}\cdot\vec{v}=Fv\cos\theta$. This is why a car engine delivering constant power can reach high speed only by reducing its driving force (lower gears give more force at low speed). The kilowatt-hour (kWh) used in electricity billing is a unit of energy, not power: $1\ \text{kWh}=3.6\times 10^6\ \text{J}$.
Collisions are brief interactions in which bodies exert large forces on each other. In every collision, linear momentum is conserved (no net external force during the short impact). Collisions are classified by what happens to kinetic energy:
- Elastic collision: both momentum and kinetic energy are conserved. Example: collisions between hard steel balls or gas molecules (ideally).
- Inelastic collision: momentum is conserved but kinetic energy is not — some is lost to heat, sound or deformation.
- Perfectly inelastic collision: the bodies stick together and move with a common velocity; the maximum possible kinetic energy is lost (consistent with momentum conservation).
Elastic collision in one dimension. For two bodies $m_1,m_2$ with initial velocities $u_1,u_2$, solving momentum and energy conservation gives the final velocities $v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}$ and $v_2=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}$. A neat consequence: in a 1D elastic collision the relative velocity of approach equals the relative velocity of separation, $u_1-u_2=v_2-v_1$. When equal masses collide elastically and one is at rest, they simply exchange velocities.
Perfectly inelastic collision. If the bodies stick together, momentum conservation gives the common velocity $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$.
Coefficient of restitution. Real collisions lie between the two extremes, described by $e=\frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}=\frac{v_2-v_1}{u_1-u_2}$. Here $e=1$ for a perfectly elastic collision, $e=0$ for a perfectly inelastic one, and $0
A pump lifts 600 kg of water through a height of 10 m in 20 s. Find the power of the pump. Take $g=10\ \text{m/s}^2$.
Solution- Step 1: Work done against gravity $W=mgh=600\times 10\times 10=60000\ \text{J}$.
- Step 2: Power $P=\frac{W}{t}=\frac{60000}{20}$.
- Step 3: $P=3000\ \text{W}=3\ \text{kW}$.
Answer: $P=3000\ \text{W}=3\ \text{kW}$
A car moves at a constant velocity of 15 m/s while the engine exerts a forward force of 800 N. Find the power developed by the engine.
Solution- Step 1: For force along the velocity, $P=\vec{F}\cdot\vec{v}=Fv\cos 0^\circ=Fv$.
- Step 2: Substitute $F=800\ \text{N}$, $v=15\ \text{m/s}$.
- Step 3: $P=800\times 15=12000\ \text{W}=12\ \text{kW}$.
Answer: $P=12000\ \text{W}=12\ \text{kW}$
A 2 kg ball moving at 6 m/s collides head-on and elastically with a stationary 2 kg ball. Find the velocities of both balls after the collision.
Solution- Step 1: The masses are equal and the second ball is at rest ($u_1=6\ \text{m/s}$, $u_2=0$, $m_1=m_2$).
- Step 2: For equal masses in an elastic collision, the velocities are simply exchanged: $v_1=u_2$ and $v_2=u_1$.
- Step 3: Therefore $v_1=0$ and $v_2=6\ \text{m/s}$.
Answer: First ball stops ($v_1=0$); second ball moves at $6\ \text{m/s}$.
A 3 kg body moving at 4 m/s collides with and sticks to a 1 kg body at rest (perfectly inelastic). Find the common velocity after the collision.
Solution- Step 1: For a perfectly inelastic collision, momentum is conserved and the bodies move together: $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$.
- Step 2: Substitute $m_1=3\ \text{kg}$, $u_1=4\ \text{m/s}$, $m_2=1\ \text{kg}$, $u_2=0$.
- Step 3: $v=\frac{3\times 4+1\times 0}{3+1}=\frac{12}{4}=3\ \text{m/s}$.
Answer: $v=3\ \text{m/s}$
In the perfectly inelastic collision of Example 4 (3 kg at 4 m/s sticking to 1 kg at rest, common velocity 3 m/s), find the kinetic energy lost.
Solution- Step 1: Initial KE $=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}\times 3\times 4^2+0=24\ \text{J}$.
- Step 2: Final KE $=\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}\times 4\times 3^2=18\ \text{J}$.
- Step 3: KE lost $=24-18=6\ \text{J}$ (converted to heat, sound and deformation).
Answer: KE lost $=6\ \text{J}$
A ball is dropped from a height of 5 m and rebounds to a height of 1.25 m. Find the coefficient of restitution between the ball and the floor.
Solution- Step 1: Speed of approach $u=\sqrt{2gh_1}$ and speed of separation $v=\sqrt{2gh_2}$, so $e=\frac{v}{u}=\sqrt{\frac{h_2}{h_1}}$.
- Step 2: Substitute $h_1=5\ \text{m}$, $h_2=1.25\ \text{m}$, giving $e=\sqrt{\frac{1.25}{5}}=\sqrt{0.25}$.
- Step 3: $e=0.5$.
Answer: $e=0.5$