01 Making Measurement

1.1 Length, Volume and Mass

1 Topic | 3 Quizzes
1.2 Density

1 Topic | 2 Quizzes
1.3 Measuring Time

1 Topic | 1 Quiz
02 Describing motion

2.1 Motion

3 Topics | 4 Quizzes
2.3 Mass and Weight

3 Topics | 4 Quizzes
2.5 Forces and Its Effects

1 Topic | 9 Quizzes
2.10 Momentum

3 Topics | 3 Quizzes
2.11 Work, Energy and Power

10 Topics | 7 Quizzes
2.12 Pressure

1 Topic | 2 Quizzes
Thermal Physics

3.1 Simple Kinetic Molecular Model of Matter

4 Topics | 5 Quizzes
3.2 Thermal Properties and Temperature

8 Topics | 4 Quizzes
3.3 Thermal Process

1 Topic | 1 Quiz
04 Waves

4.1 General Wave Properties

3 Topics | 3 Quizzes
4.2 Light

5 Topics | 5 Quizzes
4.3 Electromagnetic Spectrum

2 Quizzes
4.4 Sound

4 Quizzes
Lesson Progress

0% Complete

Question (a)(i)

Fig. 3.1 shows a skier descending a hillside. Fig. 3.2 shows the speed/time graph of his motion.

How can you tell that the acceleration of the skier is constant during the 8 s shown on the graph?

The graph is a straight line.

**Explanation:** A straight line on a speed/time graph indicates a constant acceleration.

Question (a)(ii)

Calculate the acceleration of the skier.

The acceleration is 0.75 m/s².

**Explanation: **The acceleration is calculated using \(a = \frac{Δv}{t} = \frac{6 \text{ m/s}}{8 \text{ s}} = 0.75 \text{ m/s}²\).

Question (b)(i)

Another skier starts from rest at the top of the slope. As his speed increases, the friction force on the skier increases. State the effect of this increasing friction force on the acceleration.

The acceleration decreases.

**Explanation:** Increasing friction force opposes the motion, reducing the acceleration.

Question (b)(ii)

Eventually, the speed of the skier becomes constant. What can be said about the friction force when the speed is constant?

The friction force is equal to the forward/downward force.

**Explanation:** When the speed is constant, the forces are balanced, so the friction force equals the forward/downward force.

Question (b)(iii) 1.

On the axes of Fig. 3.3, sketch a possible speed/time graph for the motion of the second skier.

The graph starts at the origin, curves from the start with a decreasing gradient, and has a horizontal final part.

**Explanation: **This shape represents the skier's acceleration decreasing over time until reaching a constant speed.

Question (b)(iii) 2.

On your graph, mark with the letter A a region where the acceleration is not constant. Mark with the letter B the region where the speed is constant.

– Label A on any correct curved region.

– Label B on horizontal region.

– Label B on horizontal region.

**Explanation: **The curved region indicates changing acceleration, and the horizontal region indicates constant speed.

Question (a)

Two students make the statements about acceleration that are given below.

Student A: For a given mass the acceleration of an object is proportional to the resultant force applied to the object.

Student B: For a given force the acceleration of an object is proportional to the mass of the object.

One statement is correct and one is incorrect.

Re-write the incorrect statement, making changes so that it is now correct.

For a given ______, the acceleration of an object is ______.

For a given force, the acceleration of an object is inversely proportional to the mass of the object.

**Explanation: **The correct relationship is that acceleration is inversely proportional to mass for a given force, according to Newton's second law of motion.

Question (b)

State the equation which links acceleration \(a\), resultant force \(F\) and mass \(m\).

The equation is \(F = ma\).

**Explanation:** This is Newton's second law of motion, stating that force equals mass times acceleration.

Question (c)(i)

Describe what happens to the motion of a moving object when there is no resultant force acting on it.

The object continues with constant speed and direction.

**Explanation: **When no resultant force acts on an object, it maintains its state of motion, according to Newton's first law of motion.

Question (c)(ii)

Describe what happens to the motion of a moving object when a resultant force is applied to it in the opposite direction to the motion.

The object slows down.

**Explanation:** A force opposite to the direction of motion causes deceleration, reducing the object's speed.

Question (c)(iii)

Describe what happens to the motion of a moving object when a resultant force is applied to it in a perpendicular direction to the motion.

The object moves in a curved path.

**Explanation:** A perpendicular force causes the object to change direction, resulting in a curved trajectory.

Question (a)

The apparatus shown in Fig. 5.1 is used to demonstrate how a coin and a piece of paper fall when they are released from rest.

At the positions shown in Fig. 5.1, the paper is descending at constant speed but the coin still accelerates.

In terms of the forces acting, explain these observations.

Paper

The paper experiences air resistance equal to its weight.

**Explanation:** The forces balance out, so there is no resultant force, resulting in no acceleration.

Coin

The coin experiences a greater weight than air resistance.

**Explanation: **The downward force is larger than the upward force of air resistance, causing the coin to continue accelerating.

Question (b)

A vacuum pump is now connected at A and the air in the tube is pumped out. The paper and coin are again made to fall from rest. State one difference that would be observed, compared with what was observed when air was present.

Both the paper and the coin will fall at the same speed and hit the bottom at the same time.

**Explanation: **In the absence of air, there is no air resistance acting on the paper, so both objects fall freely under gravity.

Question (a)(i)

A hillside is covered with snow. A skier is travelling down the hill.

The table below gives the values of the acceleration of the skier at various heights above the bottom of the hill.

\begin{align}

\begin{array}{|l|c|c|c|c|}

\hline \text { height } / \mathrm{m} & 350 & 250 & 150 & 50 \\

\hline \frac{\text { acceleration }}{\mathrm{m} / \mathrm{s}^2} & 7.4 & 3.6 & 1.2 & 0 \\

\hline

\end{array}

\end{align}

On Fig. 1.2, plot the values given in the table, using dots in circles. Draw the best curve for these points.

Plot all points correctly ±½ small square, and draw a smooth curve through the points.

**Explanation:** Ensure accurate plotting and a smooth curve that best fits the plotted points.

Question (b)(i)

Describe what is happening, during the descent, to the acceleration of the skier.

The acceleration is decreasing.

**Explanation: **As the skier descends, the acceleration reduces due to factors such as increased friction.

Question (b)(ii)

Describe what is happening, during the descent, to the speed of the skier.

The speed of the skier is increasing.

**Explanation:** Despite the decreasing acceleration, the skier continues to gain speed, though at a slower rate.

Question (c)

The acceleration becomes zero before the skier reaches the bottom of the hill. Use ideas about forces to suggest why this happens.

The resultant force becomes zero.

**Explanation:** When the forces acting on the skier balance out, there is no net force, resulting in zero acceleration.

Question (d)

Below a height of 50 m, further measurements show that the acceleration of the skier has a negative value. What does this mean is happening to the speed of the skier in the last 50 m?

The speed of the skier is decreasing.

**Explanation: **Negative acceleration indicates deceleration, meaning the skier is slowing down in the last 50 meters.

Question (e)

The skier has a mass of 60 kg. Calculate the resultant force on the skier at a height of 250 m.

The resultant force is 216 N.

**Explanation: **The force is calculated using \(F = ma\), where \(a = 3.6 \text{ m/s}²\). Hence, \(F = 60 \times 3.6 = 216 \text{ N}\).

In a laboratory, an experiment is carried out to measure the acceleration of a trolley on a horizontal table, when pulled by a horizontal force.

Fig. 1.1

The measurements are repeated for a series of different forces, with the results shown in the table below

\begin{align}

\begin{array}{|l|c|c|c|c|}

\hline \text { force } / \mathrm{N} & 4.0 & 6.0 & 10.0 & 14.0 \\

\hline \frac{\text { acceleration }}{\mathrm{m} / \mathrm{s}^2} & 0.50 & 0.85 & 1.55 & 2.25 \\

\hline

\end{array}

\end{align}

(a) On Fig. 1.2, plot these points and draw the best straight line for your points.

All points should be plotted within ±1/2 small square, with a straight line of best fit through the candidate's points.

**Explanation: **Accurate plotting of data points and drawing a best-fit line helps visualize the relationship between variables.

(b) (i) Find the value of this force.

The value of the force is 1.2 N.

**Explanation: **This value is derived from the candidate’s correct data with unit, ensuring precision in measurements.

(ii) A force smaller than that in (b)(i) is applied to the stationary trolley. Suggest what happens to the trolley, if anything.

The trolley remains stationary, with no acceleration.

**Explanation:** A force smaller than 1.2 N is insufficient to overcome friction and initiate motion in the trolley.

(c) Show that the gradient of your graph is about 5.7.

Using the data from the candidate's graph for ΔF and Δm, the gradient is calculated to be approximately 5.7.

**Explanation:** The gradient represents the ratio of force to acceleration, indicating the trolley's mass-related characteristics.

(d) (i) State the equation that links resultant force F, mass m and acceleration a.

The equation is F = ma.

**Explanation: **Newton's second law of motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

(ii) Use your gradient from (c) to find the mass of the trolley.

The mass of the trolley is approximately 0.175 kg.

**Explanation: **By rearranging F = ma to m = F/a and using the gradient, the trolley's mass is calculated.

(e) On Fig. 1.3, sketch a speed/time graph for a trolley with constant acceleration.

The graph should be a straight line of positive gradient.

**Explanation:** Constant acceleration results in a linear increase in speed over time, depicted as a straight line.

A person is standing on the top of a cliff, throwing stones into the sea below.

(a) The person throws a stone horizontally.

(i) On Fig. 2.1, draw a line to show the path which the stone might take between leaving the person’s hand and hitting the sea.

The path of the stone is a downward curve, starting horizontally at the top and not becoming vertical at the bottom.

**Explanation: **The stone follows a curved trajectory due to the influence of gravity, which causes it to accelerate downwards while maintaining its horizontal velocity.

(ii) On the line you have drawn, at a point halfway to the sea, mark the stone and the direction of the force on the stone.

The force on the stone is vertically downward.

**Explanation: **The only force acting on the stone in free fall (neglecting air resistance) is the force of gravity, which acts vertically downward.

(b) Later, the person drops a small stone and a large stone vertically from the edge of the cliff.

Comment on the times taken for the two stones to hit the water.

The times taken for the two stones to hit the water are the same because air resistance is negligible, resulting in the same acceleration.

**Explanation: **In the absence of significant air resistance, both stones experience the same gravitational acceleration, leading to identical fall times.

(c) 800 m from the point where the person is standing, a navy ship is having target practice.

The person finds that if a stone is dropped vertically at the same time as the spurt of smoke from the ship’s gun is seen, the stone hits the water at the same time as the sound from the gun is heard.

Sound travels at 320 m/s in that region.

Calculate the velocity with which the stone hits the water.

The velocity with which the stone hits the water is 25 m/s.

**Explanation: **Using the formula \( v = at \) and given that the time is calculated as 2.5 seconds, the velocity is \( v = 10 \times 2.5 = 25 \, \text{m/s} \).

Fig. 2.1 is a head-on view of an airliner flying at constant speed in a circular horizontal path. The centre of the circle is to the left of the diagram.

(a) On Fig. 2.1, draw the resultant force acting on the airliner. Explain your answer.

The resultant force is horizontal to the left. This is because the airliner is accelerating or changing direction towards the center of the circle, which indicates a centripetal force is acting towards the center.

**Explanation: **The force that acts on the airliner is towards the center of the circular path, causing the centripetal acceleration necessary to maintain the circular motion.

(b) The weight of the airliner is 1.20 × 10⁶ N and there is an aerodynamic lift force of 1.39 × 10⁶ N acting at 30° to the left of the vertical. By drawing a scale vector diagram, or otherwise, show that the resultant of these two forces is in the same direction as the resultant force you drew in (a).

Drawing a scale vector diagram or using trigonometry confirms that the resultant of the weight and lift forces points horizontally to the left, consistent with the direction of the centripetal force.

Explanation: The vector diagram shows that the sum of the weight and lift forces results in a horizontal force towards the left, aligning with the centripetal force direction.

(c) The speed is constant as the airliner flies in this circular path. State and explain what is happening to the velocity.

The direction of the velocity is changing while its magnitude remains constant.

**Explanation: **Since the airliner is moving in a circular path at a constant speed, its direction continuously changes, resulting in a change in velocity.

A brick is dropped from the top of a very tall building as it is being constructed.

Fig. 1.1 is the speed/time graph for the brick as it falls to the ground.

(a) State a time at which the acceleration of the brick is

(i) zero,

A time from 12.5 – 14.9 s or 15.1 – 16.0 s.

**Explanation:** The brick's speed is constant during these times, indicating zero acceleration.

(ii) constant but not zero,

A time from 0 – 2.5 s or 14.9 – 15.1 s.

**Explanation: **The brick's speed increases linearly, indicating constant but non-zero acceleration.

(iii) not constant.

A time from 2.5 – 12.5 s.

**Explanation:** The brick's acceleration varies as the graph shows a curve, indicating changing acceleration.

(b) Explain in terms of the forces acting on the brick why, between 0 and 14.0 s, its speed varies in the way shown by the graph.

Initially, weight/force of gravity and air friction/resistance act. It speeds up/accelerates and air friction/resistance increases. It reaches terminal/constant velocity. Air friction/resistance equals weight or no resultant force or forces in equilibrium.

**Explanation: **Gravity causes the brick to accelerate initially, but as air resistance increases, the net force decreases until it balances with gravity, resulting in terminal velocity.

(c) State the direction of the resultant force acting on the brick at time 15.0 s.

The direction of the resultant force is upwards.

**Explanation:** At 15.0 s, the brick's speed decreases, indicating an upward resultant force opposing its motion.

Fig. 2.1 shows a model fire engine used by a student to take measurements of force and motion.

The model projects a jet of water forwards. The forcemeter holds the model stationary. It indicates a force of \(0.060 \mathrm{~N}\) acting on the model.

(a) The back of the model breaks a pair of light beams and the time to pass between them is measured electronically. The beams are 12 mm apart and the second beam is broken 0.080 s after the first.

The student times with a stopwatch how long it takes from the release of the model until the beams are cut.

Calculate the time he measures.

The time measured is 5.0 s.

**Explanation: **The velocity is calculated as 0.15 m/s by dividing 12 mm by 0.080 s. The time is calculated using the equation t = (Δv)/a, which is 5.0 s after substituting the given values.

(b) This experiment is carried out with the water tank in the model nearly full.

Calculate the mass of the model including the water in the tank.

The mass of the model is 2 kg.

**Explanation:** The mass is calculated using the equation F/a = m. Substituting 0.060 N for F and 0.030 m/s² for a gives 2 kg.

(c) The student repeats the experiment with the same force but with the water tank nearly empty.

State and explain how the acceleration will compare to that of the first experiment.

The acceleration will be greater because the mass is less.

**Explanation:** With a reduced mass and the same force, the acceleration increases according to the equation F = ma.

(c) Fig. 1.3 shows a person bungee-jumping from a bridge. The person is attached to a long elastic rope.

(c)(i) In 1.5 s the speed of the jumper increases from zero to 10.5 m/s. Calculate her average acceleration during this time.

The average acceleration is 7.0 m/s².

**Explanation: **The average acceleration is calculated using the formula \(a = \frac{Δv}{t}\), where \(Δv\) is the change in velocity (10.5 m/s) and \(t\) is the time (1.5 s), resulting in an acceleration of 7.0 m/s².

(c)(ii) At one point during the fall, she reaches her maximum speed. State her acceleration at this point.

The acceleration at this point is 0 m/s².

**Explanation:** When the jumper reaches her maximum speed, her acceleration is zero because she is not speeding up or slowing down at that moment.

(c)(iii) What can be said about the forces acting on her at this point?

The upward and downward forces are equal and opposite.

**Explanation: **At maximum speed, the forces acting on the jumper, such as gravity and the tension in the rope, balance each other, resulting in no resultant force and therefore no acceleration.

A free-fall parachutist jumps from a helium balloon, but does not open his parachute for some time.

Fig. 1.1 shows the speed-time graph for his fall. Point B indicates when he opens his parachute.

(a)(i) State the value of the gradient of the graph immediately after time t = 0.

The gradient of the graph immediately after time t = 0 is 10 m/s².

**Explanation:** This value represents the acceleration due to gravity, indicating the parachutist is in freefall immediately after the jump, accelerating at approximately 10 m/s² due to gravity.

(a)(ii) Explain why the gradient has this value.

The gradient represents the parachutist's acceleration, which is due to the gravitational pull experienced during freefall, typically around 10 m/s².

(b) State how Fig. 1.1 shows that the acceleration decreased between time t = 0 and the time to A.

The graph shows a decreasing gradient from point 0 to A, indicating that the acceleration decreases over this period.

(c) Explain, in terms of forces, what is happening in section AB of the graph in Fig. 1.1.

In section AB, the speed remains constant, signifying terminal velocity. This indicates that the forces acting on the parachutist—gravity and air resistance—are balanced, resulting in no resultant force and thus no acceleration.

Fig. 2.1 shows a vehicle designed to be used on the Moon.

Fig. 2.1

The brakes of the vehicle are tested on Earth.

(a) The acceleration of free fall on the Moon is one sixth \(\left(\frac{1}{6}\right)\) of its value on Earth.

Tick one box in each column of the table to predict the value of that quantity when the vehicle is used on the Moon, compared to the test on Earth.

\begin{align}

\begin{array}{|c|c|c|c|}

\hline & \begin{array}{l}

\text { mass of vehicle on } \\

\text { Moon }

\end{array} & \begin{array}{l}

\text { weight of vehicle on } \\

\text { Moon }

\end{array} & \begin{array}{l}

\text { deceleration of vehicle } \\

\text { on Moon with same } \\

\text { braking force }

\end{array} \\

\hline 10 \times \text { value on Earth } & & & \\

\hline 6 \times \text { value on Earth } & & & \\

\hline \text { same as value on Ear } & & & \\

\hline \begin{array}{l}

\frac{1}{6} \times \text { value on Earth }

\end{array} & & & \\

\hline \begin{array}{l}

\frac{1}{10} \times \text { value on Earth }

\end{array} & & & \\

\hline

\end{array}

\end{align}

– The mass of the vehicle on the Moon is the same as on Earth.

– The weight of the vehicle on the Moon is 1/6th of its value on Earth.

– The deceleration of the vehicle on the Moon with the same braking force is the same as on Earth.

**Explanation:**– Mass is a measure of inertia and does not change with location. Thus, the mass remains the same whether on Earth or the Moon.

– Weight is a force exerted by gravity. Since the Moon's gravity is 1/6th of Earth's, the vehicle's weight is correspondingly reduced to 1/6th.

– Deceleration is influenced by the mass and the applied force. Since the mass remains constant and the braking force is the same, the deceleration does not change due to a change in gravitational acceleration alone.

– The weight of the vehicle on the Moon is 1/6th of its value on Earth.

– The deceleration of the vehicle on the Moon with the same braking force is the same as on Earth.

– Weight is a force exerted by gravity. Since the Moon's gravity is 1/6th of Earth's, the vehicle's weight is correspondingly reduced to 1/6th.

– Deceleration is influenced by the mass and the applied force. Since the mass remains constant and the braking force is the same, the deceleration does not change due to a change in gravitational acceleration alone.

Fig. 3.1 shows water flowing at very slow speed over a cliff edge.

The water falls 15 m onto the rocks below.

(a) Show that the velocity of the water when it strikes the rocks is 17 m/s.

The velocity of the water when it strikes the rocks is 17 m/s.

**Explanation: **Using the kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \) (initial velocity), \( a = 9.8 \) m/s² (acceleration due to gravity), and \( s = 15 \) m (distance fallen), we find \( v = \sqrt{2 \times 9.8 \times 15} = 17.15 \) m/s, which rounds to approximately 17 m/s.

(b) 30 kg of water flows over the cliff edge every second. Calculate the force exerted by the rocks on the falling water. Ignore any splashing.

The force exerted by the rocks on the falling water is 520 N.

**Explanation:** The force exerted can be calculated using the rate of change of momentum formula, \( F = \Delta p / \Delta t \). With \( \Delta p = m \times v = 30 \times 17.15 \) kg·m/s and \( \Delta t = 1 \) s, we calculate \( F = 514.5 \) N, which rounds to approximately 520 N.

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