Exam practice

A reference point is a fixed point, line or place (or object) with which we compare the position of an object to decide whether it is at rest or in motion, and how it is moving.

Acceleration is the rate of change of velocity of a body: the change in its velocity in one unit of time. $a = (v - u)/t$. Its SI unit is m/s².

Retardation is the rate of decrease of velocity of a body. It happens when the final velocity is less than the initial velocity, as when a vehicle brakes. It is negative acceleration. For example, a bus that slows from 20 m/s to rest in 4 s has a retardation of 5 m/s².

The relative velocity of a body is its velocity compared with another body (or a reference point): how it seems to move to an observer on that other body. For example, a bus at 15 m/s seen from a bus going the same way at 10 m/s seems to move at only 5 m/s.

Same direction: $v_{AB} = v_A - v_B$. Example: two buses at 15 m/s and 10 m/s going the same way: $v_{AB}$ = 15 − 10 = 5 m/s.

Opposite directions: $v_{AB} = v_A + v_B$. Example: the same two buses going opposite ways: $v_{AB}$ = 15 + 10 = 25 m/s.

Average velocity is the total displacement (total distance traveled in a particular direction) of a body divided by the total time taken: $\bar v = s/t$.

Average velocity is calculated when a body is in non-uniform motion, so that its velocity keeps changing.
$\bar v = \dfrac{\text{total displacement } (s)}{\text{total time taken } (t)}$
and, when the velocity changes steadily from an initial velocity $u$ to a final velocity $v$,
$\bar v = \dfrac{u + v}{2}$

$s$ = 4.2 km = 4200 m; $t$ = 7 min = 7 × 60 = 420 s.
$$\bar v = \frac{s}{t} = \frac{4200\ \text{m}}{420\ \text{s}} = 10\ \text{m/s}$$

$u$ = 0, $v$ = 30 m/s, $t$ = 12 s.
$$a = \frac{v - u}{t} = \frac{30 - 0}{12} = 2.5\ \text{m/s}^2$$

$u$ = 0, $a$ = 5 m/s², $t$ = 6 s.
From $a = (v - u)/t$: $v = u + at = 0 + 5 \times 6 =$ 30 m/s.
The velocity rises steadily, so
$$\bar v = \frac{u + v}{2} = \frac{0 + 30}{2} = 15\ \text{m/s}$$

$v$ = 0, retardation 2 m/s², so $a$ = −2 m/s²; $t$ = 4 s.
From $a = (v - u)/t$: $-2 = (0 - u)/4$, so $u = 2 \times 4 =$ 8 m/s.

Same direction: $v_{AB} = v_A - v_B$ = 60 − 50 = 10 km/h.
Opposite directions: $v_{AB} = v_A + v_B$ = 60 + 50 = 110 km/h.

Some of the input work is always used up against friction at the fulcrum (and in bending the lever), and turns into heat. So the output work is always less than the input work, and output ÷ input × 100% is less than 100%.

A wheelbarrow is a second-class lever with its fulcrum at the wheel. Moving the load towards the wheel makes the load distance shorter, while the effort distance (wheel to handles) stays the same. By $E \times \text{E.D.} = L \times \text{L.D.}$, a smaller load distance needs a smaller effort.

VR depends only on the distances, but MA depends on the real effort. Because of friction, the real effort is always a little more than it would be in a perfect machine, so the MA (load ÷ effort) is always a little less than the VR. They would be equal only in an ideal machine with no friction.

Put the fulcrum close to the load and apply the effort far from the fulcrum. A long effort distance and a short load distance mean, by $E \times \text{E.D.} = L \times \text{L.D.}$, that a small effort balances, and then lifts, a big load. For example, with the load 0.5 m from the fulcrum and the effort 1.5 m from it, an effort of 267 N lifts 800 N.

A first-class lever: the pivot (fulcrum) is between the blades, which cut the cloth or paper (load), and the handles, where the fingers press (effort).

It multiplies the effort: it can lift or move a load that is heavier than the effort applied. For example, with an MA of 4, an effort of 150 N lifts a load of 600 N.

• It can change the direction of the force: push down on a crowbar and the rock goes up.
• It can give speed and reach: a third-class lever such as a broom, a fishing rod or our forearm makes the far end move a long way, fast, for a small movement of the hand.
• It lets us work safely and conveniently: tongs keep our fingers away from hot coal, and pliers grip what fingers can’t.

Input work: input work = effort × distance moved by the effort.

$E$ = 75 N, $L$ = 300 N, L.D. = 25 cm.
$E \times \text{E.D.} = L \times \text{L.D.}$: 75 × E.D. = 300 × 25
$$\text{E.D.} = \frac{300 \times 25}{75} = 100\ \text{cm}$$

Take Nirmal as the effort and Sarad as the load: $E$ = 300 N, $L$ = 550 N, L.D. = 1.5 m.
$E \times \text{E.D.} = L \times \text{L.D.}$: 300 × E.D. = 550 × 1.5 = 825
$$\text{E.D.} = \frac{825}{300} = 2.75\ \text{m}$$
Nirmal sits 2.75 m from the center, on the other side.

L.D. = 50 cm = 0.5 m; the effort is at the other end of the 2 m lever, so E.D. = 2 − 0.5 = 1.5 m.
$$\text{MA} = \frac{L}{E} = \frac{900}{300} = 3 \qquad \text{VR} = \frac{\text{E.D.}}{\text{L.D.}} = \frac{1.5}{0.5} = 3$$
$$\eta = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{3}{3} \times 100\% = 100\%$$
On paper this lever is a perfect machine, because the problem ignores friction. A real lever would need a little more than 300 N.

The studs touch the ground over a small area, so the player’s weight makes a large pressure under each stud. The studs sink into the grass and soil and grip it, so the player doesn’t slip when running and turning.

A sharp plowshare has a very small area at its edge. By $P = F/A$, the pull of the oxen or tractor then makes a very large pressure there, so the plowshare cuts into the soil easily.

The pressure of water increases with depth ($P = hdg$), so the water pushes hardest at the bottom of the dam. A wider, thicker base is stronger, and can stand that greater pressure without breaking.

The tap on the ground floor is farther below the water level in the tank than a tap upstairs. The pressure of the water at a tap grows with this depth ($P = hdg$), so the water comes out of the ground-floor tap with greater pressure and faster, and the bucket fills sooner.

Pressure is the force acting perpendicularly on a unit area of a surface: $P = F/A$. Its SI unit is N/m² or pascal (Pa).
Applications:

• i.Knives, sickles and axes are sharpened, so their thin edge makes a large pressure and cuts easily.
• ii.Nails and drawing pins have pointed tips, so they go into wood or a wall easily.
• iii.Tractors have wide back wheels, and trucks have many wheels, to spread their weight and reduce the pressure on the ground.

Make holes at different heights in the side of a bottle full of water (Activity 6.6). Water spurts out of every hole, sideways: the water is pushing on the walls. The jet from the lowest hole is the strongest, which shows that the pressure is greatest at the greatest depth. Also, a balloon or an empty bottle pushed under water is pushed back up, and a funnel covered with rubber lowered into water pushes the liquid in a manometer (Activity 6.10).

• i.A dam is made thicker at the bottom than at the top, because the water pressure is greatest at the bottom.
• ii.A water tank is kept high on a house, so the water reaches the taps below with a good pressure.

(Also: deep-sea divers wear special suits; mercury bottles have thick bottoms.)

Atmospheric pressure is the pressure exerted by the air of the atmosphere on a unit area of the Earth’s surface, because of the weight of the air above it. At sea level it is about 760 mmHg, or about 100,000 Pa.
Uses:

• i.Filling a syringe with medicine, or a fountain pen with ink: pulling the plunger lowers the pressure inside, and the atmosphere pushes the liquid in.
• ii.Drinking through a straw, and lifting water with a hand pump.

Take a container with a base of area $A$, filled to a height $h$ with a liquid of density $d$, where the acceleration due to gravity is $g$.
Volume of the liquid: $V = A \times h$
Mass of the liquid: $m = d \times V = d \times A \times h$
Weight of the liquid (the force on the base): $W = mg = d \times A \times h \times g$
Pressure on the base:
$$P = \frac{W}{A} = \frac{d \times A \times h \times g}{A} = h\,d\,g$$
Hence proved: $P = hdg$.

Draw a U-shaped tube with a liquid (colored water) in its bend. Label: the U-tube, the liquid, the left arm joined to the gas whose pressure is measured (pressure P₂), the right arm open to the air (pressure P₁), the scale beside it, and the difference in levels, h. If P₂ is greater than P₁, the liquid must be drawn lower in the left arm and higher in the right, as in Figure 6.21.

Atmospheric pressure is the pressure exerted by the air of the atmosphere on a unit area of the Earth’s surface. Air has weight; the whole layer of air above us presses down on everything, with about 100,000 Pa (760 mmHg) at sea level, and less at higher places.
Examples that show it:

• i.A glass full of water, covered with a card and turned upside down, doesn’t spill: the air pushes up on the card (Activity 6.7).
• ii.A can with a little boiling water, closed and then cooled, is crushed: the steam inside condenses, the pressure inside falls, and the air outside crushes the can (Activity 6.8).
• iii.A syringe fills, and juice rises up a straw, because the air pushes the liquid in when the pressure inside is lowered.

The crushed can (Activity 6.8): put a little water in a tin can with a tight lid and boil it until steam comes out. Turn off the heat, close the lid tightly at once, and pour cold water over the can. The can is crushed with a loud noise.
Why: while the water boils, the steam drives most of the air out of the can. Cooling turns the steam into a few drops of water, so the pressure inside becomes much lower than outside. The pressure of the air outside crushes the can, which shows that the atmosphere has pressure.
(The glass of water covered with a card and turned upside down, Activity 6.7, is also a good answer.)

• i.Filling a syringe with medicine: pulling the plunger lowers the pressure inside the syringe, and the air pressing on the medicine’s surface pushes it in.
• ii.Lifting water with a hand pump (and drinking through a straw): pulling the piston lowers the pressure inside the pipe, and the air pressing on the water in the well pushes it up.

Pressure: $P = F/A$. Its SI unit is N/m², or the pascal (Pa).

Atmospheric pressure. At sea level it is about 760 mmHg (about 100,000 Pa), and it is measured with a barometer.

From $P = hdg$, at the same depth the pressure of a liquid is proportional to its density: the denser the liquid, the greater the pressure. For example, mercury is 13.6 times as dense as water, so at the same depth its pressure is 13.6 times as much.

A manometer (such as a U-tube manometer): the difference $h$ between the liquid levels in its two arms measures the pressure.

Given: force (weight) $F$ = 750 N, area $A$ = 25 m².
$$P = \frac{F}{A} = \frac{750\ \text{N}}{25\ \text{m}^2} = 30\ \text{N/m}^2$$
The box puts 30 N/m² (30 Pa) on the ground.

Given: $h$ = 8 m, $d$ = 1000 kg/m³, $g$ = 9.8 m/s².
$$P = h\,d\,g = 8 \times 1000 \times 9.8 = 78{,}400\ \text{Pa}$$

Given: $h$ = 2 m, $P$ = 24,696 N/m², $g$ = 9.8 m/s². To find: $d$.
From $P = h\,d\,g$:
$$d = \frac{P}{h\,g} = \frac{24{,}696}{2 \times 9.8} = \frac{24{,}696}{19.6} = 1260\ \text{kg/m}^3$$
The liquid’s density is 1260 kg/m³: denser than water, about the same as glycerine.
Some books print this problem with 500 N/m², which gives 25.5 kg/m³, lighter than any liquid; the method is the same.