Edexcel iGCSE Physics (4PH1) 1.3 Forces, Movement & Changing Shape Exam Style Question Paper 2B - New Syllabus
Question
This question is about gravitational field strength.
(a)(i) The table shows gravitational field strengths at the surfaces of four different planets.
The planets have the same mass but different diameters.
| Diameter in \(\mathrm{km}\) | Gravitational field strength in \(\mathrm{N\,kg^{-1}}\) |
|---|---|
| 13000 | 9.8 |
| 18000 | 5.1 |
| 21000 | 3.8 |
| 24000 | 2.9 |
Give a reason why a scatter graph is the correct way to display this data, rather than a bar chart. (1)
(a)(ii) A student claims that the relationship between the gravitational field strength and the diameter of a planet is given by this formula.
\(\mathrm{gravitational\ field\ strength}\times(\mathrm{diameter})^2=\mathrm{constant}\)
Using data from the table, evaluate the student’s claim. (4)
(b) The diameter of a planet affects the gravitational field strength at the surface of the planet.
Which of these is also a factor that affects gravitational field strength? (1)
A. absolute magnitude
B. colour
C. mass
D. specific heat capacity
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
▶️ Answer/Explanation
(a)(i) Why a scatter graph is appropriate [1 mark]
Both diameter and gravitational field strength are continuous variables. Therefore, a scatter graph is appropriate for showing the relationship between the two variables.
Final Answer: \( \boxed{\mathrm{Both\ variables\ are\ continuous}} \)
(a)(ii) Evaluating the student’s claim [4 marks]
1. Use one row of data to calculate the constant:
\(\mathrm{constant}=(9.8)(13000)^2\)
\(\mathrm{constant}=1\,656\,200\,000\)
2. Use a second row of data:
\(\mathrm{constant}=(5.1)(18000)^2\)
\(\mathrm{constant}=1\,652\,400\,000\)
3. Compare the values:
The two calculated constants are very close:
\(1.6562\times10^9\) and \(1.6524\times10^9\)
4. Conclusion:
The values are sufficiently close to be considered approximately constant. Therefore, the student’s claim is supported by the data.
For example, the other two rows give:
\((3.8)(21000)^2=1\,675\,800\,000\)
\((2.9)(24000)^2=1\,670\,400\,000\)
These values are also close to the other calculated constants.
Final Answer: \( \boxed{\mathrm{The\ student’s\ claim\ is\ supported}} \), because the calculated values of \(\mathrm{gravitational\ field\ strength}\times(\mathrm{diameter})^2\) are approximately constant.
(b) Factor affecting gravitational field strength [1 mark]
The gravitational field strength at the surface of a planet depends on the mass of the planet as well as its size.
Correct Answer: \( \boxed{\mathrm{C.\ mass}} \)
Final Answer: C. mass
Question
A tennis ball is dropped from a height above the ground.
(a) Diagram 1 shows the tennis ball at the instant it is released.
On diagram 1, draw a labelled arrow to show the force acting on the tennis ball at the instant it is released.
(b) Diagram 2 shows the tennis ball at the instant it hits the ground.
Just before hitting the ground, the tennis ball is moving downwards at a velocity of \(2.5\,\mathrm{m\,s^{-1}}\).
Just after hitting the ground, the tennis ball is moving upwards at a speed of \(1.9\,\mathrm{m\,s^{-1}}\).
(i) The mass of the tennis ball is \(0.057\,\mathrm{kg}\).
Calculate the change in momentum of the tennis ball when it hits the ground.
Give the unit. (4)
change in momentum = __________________ unit = __________________
(ii) The ball is in contact with the ground for a time of \(6.0\,\mathrm{ms}\).
Calculate the force exerted on the ball by the ground.
Give the direction of this force. (3)
force = __________________ \(\mathrm{N}\)
direction = __________________
(c) The coefficient of restitution can be used to determine how high a ball will bounce. The higher the value of the coefficient of restitution, the higher the ball will bounce.
Diagram 3 shows a ball being dropped from a height \((h_1)\) above the ground.
The ball hits the ground and bounces.
At the top of its bounce, the ball reaches a height \((h_2)\) above the ground.
The coefficient of restitution can be determined using the formula
\(\mathrm{coefficient\ of\ restitution}=\sqrt{\dfrac{h_2}{h_1}}\)
A student decides to do an investigation to compare the coefficient of restitution for different types of ball.
Design a method for the student’s investigation.
Your answer should include details of:
- the variables in the investigation
- how to obtain accurate measurements
- how to obtain reliable results
You may add to diagram 3 or draw an additional diagram to help your answer. (6)
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.25P: Momentum, \(p=mv\), part (b)(i)
• 1.26P: Change in momentum, part (b)(i)
• 1.28P: Force, change in momentum and time, part (b)(ii)
• 1.5: Investigating motion experimentally, including identifying variables, making accurate measurements and obtaining reliable results, part (c)
▶️ Answer/Explanation
(a) Force acting on the ball [2 marks]
The force acting on the ball when it is released is its weight.
- Draw an arrow vertically downwards through the centre of the ball.
- Label the arrow weight or gravitational force.
Final Answer: A downward arrow through the centre of the ball labelled weight.
(b)(i) Change in momentum [4 marks]
1. Calculate the initial momentum:
\(p=mv\)
\(p_\mathrm{initial}=(0.057)(2.5)\)
\(p_\mathrm{initial}=0.1425\,\mathrm{kg\,m\,s^{-1}}\)
2. Calculate the final momentum:
\(p_\mathrm{final}=(0.057)(1.9)\)
\(p_\mathrm{final}=0.1083\,\mathrm{kg\,m\,s^{-1}}\)
The two velocities are in opposite directions, so the change in momentum is the sum of the two momentum magnitudes.
\(\Delta p=0.1425+0.1083\)
\(\Delta p=0.2508\,\mathrm{kg\,m\,s^{-1}}\)
Final Answer: \(\boxed{0.25\,\mathrm{kg\,m\,s^{-1}}}\)
(b)(ii) Force exerted by the ground [3 marks]
1. Use the relationship between force, change in momentum and time:
\(F=\dfrac{\Delta p}{\Delta t}\)
2. Convert the contact time:
\(6.0\,\mathrm{ms}=0.0060\,\mathrm{s}\)
3. Substitute:
\(F=\dfrac{0.25}{0.0060}\)
\(F\approx41.7\,\mathrm{N}\)
The ground pushes the ball upwards, causing the ball to reverse its direction of motion.
Final Answer: \(\boxed{42\,\mathrm{N}}\), upwards
(c) Investigation of coefficient of restitution [6 marks]
Variables:
- Independent variable: type of ball.
- Dependent variable: height of the bounce, \(h_2\), or the calculated coefficient of restitution.
- Control variables: initial drop height \(h_1\), surface on which the ball bounces, and conditions of release.
Method for accurate measurements:
- Use a metre rule or tape measure positioned vertically and perpendicular to the ground.
- Use a set square to help ensure that the measurements are vertical.
- Release each ball from the same measured height \(h_1\), without pushing it.
- Measure the maximum height \(h_2\) reached by the ball after the bounce.
- A slow-motion camera can be used to identify the maximum height accurately.
Calculation:
For each ball, calculate
\(\mathrm{coefficient\ of\ restitution}=\sqrt{\dfrac{h_2}{h_1}}\)
Reliable results:
- Repeat the experiment several times for each type of ball.
- Identify any anomalous results and repeat the measurement if necessary.
- Calculate a mean value for the coefficient of restitution for each type of ball.
Final Answer: Compare the mean coefficient of restitution values for the different balls. The ball with the highest mean coefficient of restitution has the greatest bounce.
Question
The diagram shows the collision between two balls, A and B. The masses and velocities of both balls are shown before and after the collision. Ball B is stationary before the collision.
(a) When the balls collide, ball B applies a force on ball A, which causes the velocity of ball A to change. Ball A also applies a force on ball B during the collision. Describe how the force applied on ball A compares with the force applied on ball B during the collision.
(b) Calculate the momentum of ball A before the collision.
(c) Show that the velocity, \(v\), of ball B after the collision is about \(0.6\,\mathrm{m\,s^{-1}}\).
(d) A collision is considered elastic if the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Using data from the diagram, deduce whether this collision is elastic.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.28P: Force, Momentum, and Time — part (a)
• 1.29P: Newton’s Third Law — part (a)
• 1.13–1.15: Scalars, Vectors, and Force as a Vector Quantity; Resultant Forces — part (a)
• 4.14: Kinetic Energy — part (d)
• 4.15: Conservation of Energy in Mechanical Systems — part (d)
▶️ Answer/Explanation
(a) Forces during the collision [2 marks]
- The force applied by ball A on ball B is equal in magnitude to the force applied by ball B on ball A.
- The two forces act in opposite directions.
This is an example of Newton’s Third Law. The forces form an action-reaction pair and act on different balls.
(b) Momentum of ball A before the collision [2 marks]
Use the momentum equation:
\(p=m v\)
From the diagram:
\(m=0.018\,\mathrm{kg}\)
\(v=4.9\,\mathrm{m\,s^{-1}}\)
Therefore:
\(p=0.018\times4.9\)
\(p=0.0882\,\mathrm{kg\,m\,s^{-1}}\)
So the momentum of ball A is approximately:
\(\boxed{0.088\,\mathrm{kg\,m\,s^{-1}}}\)
(c) Velocity of ball B after the collision [4 marks]
Momentum is conserved in the collision.
Total momentum before = total momentum after
Ball B is initially stationary, so its initial momentum is zero.
Therefore:
\(0.088=0.018\times(-3.5)+0.265v\)
Rearranging:
\(0.088+0.063=0.265v\)
\(v=\dfrac{0.151}{0.265}\)
\(v\approx0.57\,\mathrm{m\,s^{-1}}\)
Therefore:
\(\boxed{v\approx0.6\,\mathrm{m\,s^{-1}}}\)
(d) Determining whether the collision is elastic [4 marks]
For an elastic collision, the total kinetic energy before the collision must equal the total kinetic energy after the collision.
Use:
\(E_{\mathrm{k}}=\dfrac{1}{2}mv^2\)
Kinetic energy before the collision:
Ball B is stationary, so its initial kinetic energy is zero.
\(E_{\mathrm{k,before}}=\dfrac{1}{2}(0.018)(4.9)^2\)
\(E_{\mathrm{k,before}}\approx0.216\,\mathrm{J}\)
Kinetic energy after the collision:
For ball A:
\(E_{\mathrm{k,A}}=\dfrac{1}{2}(0.018)(3.5)^2\approx0.110\,\mathrm{J}\)
For ball B:
\(E_{\mathrm{k,B}}=\dfrac{1}{2}(0.265)(0.57)^2\approx0.043\,\mathrm{J}\)
Therefore:
\(E_{\mathrm{k,after}}\approx0.110+0.043=0.153\,\mathrm{J}\)
Since \(0.216\,\mathrm{J}\neq0.153\,\mathrm{J}\), the total kinetic energy is not conserved.
Therefore, the collision is not elastic. It is an inelastic collision.
Question
(a) The boxes show some physical quantities and their units. Draw a straight line from each physical quantity to its correct unit. One has been done for you.
(b) Some physical quantities are scalars and other physical quantities are vectors.
(i) State the difference between a scalar quantity and a vector quantity.
(ii) Give an example of a scalar quantity.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.2: Scalars and Vectors — part (b)
▶️ Answer/Explanation
(a) Physical quantities and units
Draw each line to connect the physical quantity to its correct unit, as shown in the completed mark-scheme diagram.
(b)(i) Scalars and vectors
A scalar quantity has magnitude only, whereas a vector quantity has both magnitude and direction.
(b)(ii) Example of a scalar quantity
Any correct scalar quantity is acceptable, for example mass, time, distance, or speed.
Question
The diagram shows two birds, just before bird X catches the smaller bird Y. Both birds are travelling horizontally at constant velocity.
(a) Show that the momentum of bird X is about \(5\,\mathrm{kg\,m\,s^{-1}}\).
(b) The momentum of bird Y just before it is caught is \(0.15\,\mathrm{kg\,m\,s^{-1}}\). Calculate the total momentum of the birds just before bird X catches bird Y.
(c) State the total momentum of the birds just after bird X has caught bird Y.
(d) Bird Y has a mass of \(0.17\,\mathrm{kg}\). Calculate the velocity of the birds just after bird X has caught bird Y.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.27P: Conservation of Momentum — parts (b), (c), and (d)
• 1.17–1.18: Force, Mass, Weight, and Gravitational Field Strength — part (d), for total mass
▶️ Answer/Explanation
(a) Momentum of bird X [2 marks]
Use the momentum equation:
\(p=mv\)
From the diagram, the mass of bird X is \(0.41\,\mathrm{kg}\) and its velocity is \(13\,\mathrm{m\,s^{-1}}\).
\(p=0.41\times13\)
\(p=5.33\,\mathrm{kg\,m\,s^{-1}}\)
Therefore:
\(\boxed{p\approx5\,\mathrm{kg\,m\,s^{-1}}}\)
(b) Total momentum before the collision [1 mark]
The momentum of bird X is approximately \(5.33\,\mathrm{kg\,m\,s^{-1}}\), and the momentum of bird Y is \(0.15\,\mathrm{kg\,m\,s^{-1}}\).
Since both birds are travelling in the same direction, their momenta are added:
\(p_\mathrm{total}=5.33+0.15\)
\(p_\mathrm{total}=5.48\,\mathrm{kg\,m\,s^{-1}}\)
\(\boxed{p_\mathrm{total}=5.48\,\mathrm{kg\,m\,s^{-1}}}\)
(c) Total momentum after the collision [1 mark]
Momentum is conserved provided there is no resultant external force acting on the system.
Therefore, the total momentum after bird X catches bird Y is equal to the total momentum before the collision.
\(\boxed{p_\mathrm{total}=5.48\,\mathrm{kg\,m\,s^{-1}}}\)
(d) Velocity after the collision [3 marks]
After bird X catches bird Y, the two birds move together, so their combined mass is:
\(m_\mathrm{total}=0.41+0.17\)
\(m_\mathrm{total}=0.58\,\mathrm{kg}\)
Using \(p=mv\):
\(5.48=0.58v\)
Rearranging:
\(v=\dfrac{5.48}{0.58}\)
\(v=9.45\,\mathrm{m\,s^{-1}}\)
Therefore:
\(\boxed{v\approx9.4\,\mathrm{m\,s^{-1}}}\)
The velocity is in the same direction as the original motion of the birds because the total momentum was in that direction before the collision.
Question
Diagram 1 shows a wooden plank balanced horizontally on two supports, A and B. A block is suspended from the plank between the supports by a cable of negligible weight.
(a) The weight of the block is \(260\,\mathrm{N}\).
(i) State the formula linking moment, force and perpendicular distance from the pivot.
(ii) By taking moments about support A, calculate force \(F\). Assume the weight of the plank is negligible.
(iii) Explain what will happen to the magnitude of force \(F\) if the block is moved towards support B.
(b) Diagram 2 shows the block and the cable connecting the block to the plank.
(i) The centre of gravity of the block is located at point X. Draw an arrow on diagram 2 to show the weight of the block.
(ii) The block also experiences a force due to the tension in the cable. Explain why the block remains stationary when it is supported by this tension force.
(iii) Explain why the forces acting on the block are not an example of Newton’s third law of motion.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.31P: Centre of Gravity — part (b)(i)
• 1.29P: Newton’s Third Law — part (b)(iii)
▶️ Answer/Explanation
(a)(i) Moment formula [1 mark]
The formula linking moment, force and perpendicular distance is:
\(\boxed{\mathrm{moment}=\mathrm{force}\times\mathrm{perpendicular\ distance}}\)
(a)(ii) Calculating force \(F\) [3 marks]
Take moments about support A. Since the plank is balanced, the clockwise moment equals the anticlockwise moment.
The moment of the block’s weight about A is:
\(\mathrm{moment}=260\times0.25\)
\(\mathrm{moment}=65\,\mathrm{N\,m}\)
Therefore:
\(65=F\times0.80\)
Rearranging:
\(F=\dfrac{65}{0.80}\)
\(F=81.25\,\mathrm{N}\)
\(\boxed{F\approx81\,\mathrm{N}}\)
(a)(iii) Moving the block towards support B [3 marks]
The magnitude of force \(F\) increases.
- The distance of the block from support A increases.
- Therefore, the clockwise moment produced by the block’s weight increases.
- For the plank to remain balanced, the anticlockwise moment produced by \(F\) must also increase.
Since the distance from A to the point where \(F\) acts is fixed, \(F\) must increase to produce the larger balancing moment.
(b)(i) Weight of the block [2 marks]
Draw a vertical downward arrow starting at point X, the centre of gravity of the block.
The weight acts vertically downwards through the centre of gravity.
(b)(ii) Why the block remains stationary [2 marks]
- The tension in the cable acts upwards.
- The tension is equal in magnitude to the weight of the block.
- Therefore, the forces balance and there is no resultant force.
- With no resultant force, the block has no acceleration and remains stationary.
(b)(iii) Newton’s third law [2 marks]
The tension and weight acting on the block are not a Newton’s third-law pair because both forces act on the same body, the block.
Newton’s third-law force pairs act on different bodies and are equal in magnitude and opposite in direction.
For example, the force exerted by the cable on the block has a third-law partner: the force exerted by the block on the cable.
Question
This question is about momentum and forces.
(a) State the principle of conservation of momentum.
(b) The diagram shows an air track that can be used to investigate motion without friction. Air comes out through a series of small holes in the air track. The air lifts the glider slightly above the track. A small spacecraft engine floats at rest on a cushion of air.
(i) State the momentum of the spacecraft engine when it is at rest.
(ii) The spacecraft engine ejects large numbers of xenon ions to the left. A mass of \(2.6\times10^{-8}\,\mathrm{kg}\) of xenon ions leaves the engine with a mean speed of \(26\,\mathrm{km\,s^{-1}}\). Calculate the momentum of all the ejected xenon ions.
(iii) State the magnitude and direction of the spacecraft engine’s momentum after these xenon ions leave the engine.
(iv) The ions exert a force of \(2.6\,\mathrm{mN}\) on the spacecraft engine. The spacecraft engine has a mass of \(1.2\,\mathrm{kg}\). Calculate the acceleration of the engine. Give your answer to 2 significant figures.
(c) The engine is designed to accelerate a spacecraft while the spacecraft is travelling through space. The spacecraft carries a mass of \(0.75\,\mathrm{kg}\) of xenon ions for the engine. When the engine is used, \(9.9\times10^{-8}\,\mathrm{kg}\) of xenon ions leave the engine each second. A student suggests that this small spacecraft engine would not be useful because the acceleration it produces is very small. Evaluate the student’s suggestion.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.27P: Conservation of Momentum — parts (a) and (b)(iii)
• 1.28P: Force, Momentum, and Time — parts (b)(iv) and (c)
• 1.17–1.18: Force, Mass, Weight, and Gravitational Field Strength — part (b)(iv)
▶️ Answer/Explanation
(a) Conservation of momentum [1 mark]
The total momentum of a system remains constant, provided there is no resultant external force acting on the system.
Therefore:
\(\boxed{\text{total momentum before}=\text{total momentum after}}\)
(b)(i) Momentum when at rest [1 mark]
Momentum is given by:
\(p=mv\)
The spacecraft engine is at rest, so \(v=0\).
Therefore:
\(\boxed{p=0\,\mathrm{kg\,m\,s^{-1}}}\)
(b)(ii) Momentum of the xenon ions [3 marks]
Use:
\(p=mv\)
Convert the speed into \(\mathrm{m\,s^{-1}}\):
\(26\,\mathrm{km\,s^{-1}}=26000\,\mathrm{m\,s^{-1}}\)
Substitute:
\(p=(2.6\times10^{-8})(26000)\)
\(p=6.76\times10^{-4}\,\mathrm{kg\,m\,s^{-1}}\)
Therefore:
\(\boxed{p\approx6.8\times10^{-4}\,\mathrm{kg\,m\,s^{-1}}}\)
(b)(iii) Momentum of the spacecraft engine [2 marks]
Initially, the total momentum is zero. Therefore, by conservation of momentum, the momentum of the engine must be equal in magnitude and opposite in direction to the momentum of the ejected ions.
The ions move to the left, so the spacecraft engine moves to the right.
Therefore:
\(\boxed{p=6.8\times10^{-4}\,\mathrm{kg\,m\,s^{-1}}\text{ to the right}}\)
(b)(iv) Acceleration of the engine [4 marks]
Use Newton’s second law:
\(F=ma\)
Convert the force into newtons:
\(2.6\,\mathrm{mN}=2.6\times10^{-3}\,\mathrm{N}\)
Rearrange:
\(a=\dfrac{F}{m}\)
\(a=\dfrac{2.6\times10^{-3}}{1.2}\)
\(a=2.16\times10^{-3}\,\mathrm{m\,s^{-2}}\)
To 2 significant figures:
\(\boxed{a=2.2\times10^{-3}\,\mathrm{m\,s^{-2}}}\)
(c) Evaluating the student’s suggestion [6 marks]
The student is not necessarily correct. Although the acceleration at any instant is small, the engine can operate for a long time.
The mass of xenon carried is \(0.75\,\mathrm{kg}\), while \(9.9\times10^{-8}\,\mathrm{kg}\) is used each second.
The approximate operating time is:
\(\displaystyle t=\frac{0.75}{9.9\times10^{-8}}\)
\(t\approx7.6\times10^6\,\mathrm{s}\)
This is a very long operating time. A small acceleration acting continuously for a long period can produce a significant change in velocity.
Using:
\(\Delta v=a t\)
shows that even a small acceleration can produce a large change in velocity when it acts for a sufficiently long time.
Therefore, the engine can still be useful for gradually accelerating a spacecraft over a long period.
Conclusion: the student’s suggestion is not justified.