Power and Collisions: Quick Revision

A brief note on Power and Collisions. Read it before a test. Try each question at the end before you open its answer.

The big ideas

• Power: $P = \dfrac{W}{t}$, in watts. For a force at speed $v$: $P = Fv$.
• Efficiency $=$ useful power out $\div$ power in. $1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}$ is energy.
• Collisions: momentum is always conserved (with no outside force). Kinetic energy is conserved only in an elastic collision.
• Restitution: $e = \dfrac{\text{speed of separation}}{\text{speed of approach}}$; $e = 1$ elastic, $e = 0$ sticking.
• Two dimensions: conserve momentum along each axis.

The facts at a glance

• Elastic, head-on, target at rest: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u$, $v_2 = \dfrac{2m_1}{m_1 + m_2}u$
• Equal masses, elastic: they swap velocities
• Sticking, target at rest: fraction of $K$ lost $= \dfrac{m_2}{m_1 + m_2}$
• Bouncing on a floor: each height is $e^2$ of the last
• Ballistic pendulum: momentum for the hit, energy for the swing
• Equal masses, elastic, glancing: paths at $90^\circ$

A worked example

A $2.0\ \text{kg}$ trolley at $3.0\ \text{m/s}$ meets a $1.0\ \text{kg}$ trolley coming the other way at $2.0\ \text{m/s}$, and they stick.

1. Momentum. $2.0\ \text{kg} \times 3.0\ \text{m/s} + 1.0\ \text{kg} \times (-2.0\ \text{m/s}) = 4.0\ \text{kg m/s}$.
2. Velocity after. $\dfrac{4.0\ \text{kg m/s}}{3.0\ \text{kg}} = 1.33\ \text{m/s}$, in the first trolley's direction.
3. Energy lost. Before: $9.0\ \text{J} + 2.0\ \text{J} = 11\ \text{J}$. After: $\tfrac12 \times 3.0\ \text{kg} \times (1.33\ \text{m/s})^2 = 2.67\ \text{J}$. Lost: $8.33\ \text{J}$.

Practice with new numbers

A program to try

The program collides the same two trolleys with four different coefficients of restitution.

Mistakes to avoid

• $P = \dfrac{F}{v}$. Power is force times speed.
• Kilowatt-hours as power. They are energy.
• Conserving kinetic energy in every collision. Only in elastic ones.
• Forgetting signs for bodies moving the other way.
• Energy for the hit in a ballistic pendulum. Use momentum there.

Check yourself

1. A crane lifts $2.0\ \text{t}$ through $15\ \text{m}$ in $40\ \text{s}$. What power does it give? ($g = 9.8\ \text{m/s}^2$.)

2. A cyclist delivers $250\ \text{W}$ against a steady $25\ \text{N}$ of drag. How fast does she go?

3. A $1.0\ \text{kg}$ ball at $4.0\ \text{m/s}$ hits a $3.0\ \text{kg}$ ball at rest, head-on and elastically. What are the velocities afterwards?

4. A ball dropped from $2.0\ \text{m}$ rises to $1.28\ \text{m}$. What is $e$?

5. In a ballistic pendulum, why is the block's swing so small compared with the bullet's energy?