A car of mass $1\,200$ kg hits a tree at $20$ m/s ($\sim 72$ km/h). Compare the average force on the driver if (a) the car stops in $0.05$ s (rigid body, no crumple) and (b) $0.5$ s (with crumple zone). Assume the driver decelerates with the car.

1. Acceleration $a = \dfrac{\Delta v}{\Delta t}$. Driver mass $\approx 80$ kg.
2. (a) Rigid: $a = \dfrac{0 - 20}{0.05} = -400 \text{ m/s}^2$. Force $F = ma = 80 \times 400 = 32\,000 \text{ N}$. Equivalent to roughly $40$ times the driver’s weight.
3. (b) Crumple zone: $a = \dfrac{0 - 20}{0.5} = -40 \text{ m/s}^2$. Force $F = 80 \times 40 = 3\,200 \text{ N}$. About $4$ times the driver’s weight — survivable with a seatbelt.

Key idea: extending the stopping time by $10\times$ reduces the peak force by $10\times$. Safety engineering is largely about making deceleration gentler.

A bus accelerates and then brakes sharply. Why does a standing passenger fall backwards when the bus accelerates and forwards when it brakes?

1. When the bus accelerates, the floor pushes the passenger forward through friction on their shoes. The upper body, with no such force, tends to stay in place by inertia — so the upper body moves backwards relative to the feet.
2. When the bus brakes, friction decelerates the feet while the upper body continues forward by inertia.
3. The passenger is not “thrown” — they simply continue in their previous state of motion while the bus changes its own.

A shopping trolley of mass $15$ kg is pushed with a net force of $60$ N. Find its acceleration.

1. Rearrange $F = ma$: $a = \dfrac{F}{m}$.
2. $a = \dfrac{60}{15} = 4 \text{ m/s}^2$.

A school bag has a mass of $7$ kg. Find its weight.

1. $W = mg = 7 \times 9.8 = 68.6 \text{ N}$.
2. Notice: mass is “how much stuff,” measured in kg; weight is the gravitational force on that mass, measured in newtons.

A $50$ kg box is pulled along the floor with a force of $200$ N. Friction resists with a force of $80$ N. Find the acceleration of the box.

1. Draw a free-body diagram: pull $200$ N right; friction $80$ N left; normal force and weight vertically (balance each other).
2. Net horizontal force: $F_{\text{net}} = 200 - 80 = 120 \text{ N}$.
3. $a = \dfrac{F_{\text{net}}}{m} = \dfrac{120}{50} = 2.4 \text{ m/s}^2$.

Explain how a rocket accelerates in the vacuum of space using Newton’s third law.

1. The rocket engine burns fuel and pushes hot gas out the back at high speed.
2. By Newton’s third law, the gas pushes the rocket forward with an equal and opposite force.
3. The forces act on different bodies: the rocket pushes the gas backwards; the gas pushes the rocket forwards. Both forces are equal in size.

Key idea: rockets do not push against air. They push against their own expelled exhaust, which is why they work in vacuum.

A book rests on a table. The gravitational force on the book is $10$ N downward. The normal force from the table on the book is $10$ N upward. Are these an action-reaction pair?

1. They are equal and opposite, but they act on the same object (the book).
2. Action-reaction pairs act on different objects.
3. The true pair of the book’s weight is the gravitational pull of the book on Earth ($10$ N upward on the Earth).
4. The true pair of the normal force is the book pushing down on the table ($10$ N downward on the table).

Key idea: the “equal and opposite” part is not enough to identify a third-law pair — they must act on two different objects.

A person of mass $60$ kg stands on a bathroom scale in a lift. What does the scale read (in newtons) when the lift (a) is stationary, (b) accelerates upward at $2$ m/s$^2$, (c) accelerates downward at $2$ m/s$^2$?

1. Weight $W = mg = 60 \times 9.8 = 588 \text{ N}$.
2. (a) Stationary: $F_{\text{net}} = 0$, so scale reads $588$ N.
3. (b) Upward acceleration: net force upward is $ma = 60 \times 2 = 120$ N. Scale force (up) $-$ weight (down) $= 120$. Scale $= 588 + 120 = 708$ N. You feel heavier.
4. (c) Downward acceleration: net force downward is $120$ N. Weight $-$ scale $= 120$. Scale $= 588 - 120 = 468$ N. You feel lighter.

Key idea: apparent weight is the normal force from the floor, which changes with acceleration.