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A-Level CAIE Physics 12 2 Centripetal Acceleration Question Bank

Practice A-Level CAIE Physics 12 2 Centripetal Acceleration questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

10 matching questions · Open interactive library

Question 1

1

6 marks

Question 1(b)

1(b)

A circular metal disc spins horizontally about a vertical axis, as shown in Fig. 1.1. A piece of modelling clay is attached to the disc. For the instant when the piece of modelling clay is in the position shown, draw on Fig. 1.1:

structured1 marks

Question 1(b)(ii)

1(b)(ii)

an arrow, labelled A , showing the direction of the acceleration of the modelling clay.

Mediumstructured1 marks

Answer

arrow, labelled A, pointing in NW direction B1

Question 1(c)

1(c)

The metal disc in Fig. 1.1 has a radius of 9.3 cm . The centre of gravity of the modelling clay is 1.2 cm from the rim of the disc and moves with a speed of \(0.68 \mathrm{~ms}^{-1}\).

structured2 marks

Question 1(c)(ii)

1(c)(ii)

Calculate the acceleration a of the centre of gravity of the modelling clay. a= \(\mathrm{ms}^{-2}\)

Mediumstructured2 marks

Answer

\(a=v^{2} / r\) or \(a=r \omega^{2}\) C1 \[ \begin{aligned} a =0.68^{2} /(0.093-0.012) \text { or }(0.093-0.012) \times 8.4^{2} =5.7 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 1(d)

1(d)

A second piece of modelling clay is attached to the disc in the position shown in Fig. 1.2. The second piece of modelling clay has a larger mass than the first piece. By placing one tick \((\checkmark)\) in each row, complete Table 1.1 to show how the quantities indicated compare for the two pieces of modelling clay.

Hardstructured3 marks

Answer

angular speed: same for both pieces B1 linear speed: less for second piece than first piece B1 acceleration: less for second piece than first piece B1

Question 1

1

4 marks

Question 1(c)

1(c)

During a time interval of 1400 s , the centre of gravity of the piece of modelling clay in Fig. 1.1 moves through a total distance of 0.44 m .

structured2 marks

Question 1(c)(iii)

1(c)(iii)

Calculate the magnitude of the centripetal acceleration of the piece of modelling clay. centripetal acceleration = \(\mathrm{ms}^{-2}\)

Mediumstructured2 marks

Answer

\(a=r \omega^{2}\) C1 \[ \begin{aligned} =0.18 \times\left(1.745 \times 10^{-3}\right)^{2} =5.5 \times 10^{-7} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 1(d)

1(d)

Use your answer in (c)(iii) to explain why the variation with time of the magnitude of the force exerted by the minute hand on the piece of modelling clay is negligible as the minute hand undergoes one full revolution.

Mediumstructured2 marks

Answer

centripetal acceleration is negligible compared with acceleration of free fall or numerical comparison establishing answer to (c)(iii) << 9.81 B1 resultant force is negligible compared with weight (of modelling clay) (so variation is negligible) or force exerted by minute hand (approximately) equal (and opposite) to weight of modelling clay B1

Question 1

1

5 marks

Question 1(c)

1(c)

An object rests on the surface of the Earth at the Equator. The radius of the Earth is \(6.4 \times 10^{6} \mathrm{~m}\).

structured5 marks

Question 1(c)(i)

1(c)(i)

Determine the centripetal acceleration of the object.

Mediumstructured3 marks

Answer

T=24 hours C1 \[ a=r \omega^{2} \text { and } \omega=2 \pi / T \] or \[ a=v^{2} / r \text { and } v=2 \pi r / T \] or \[ a=4 \pi^{2} r / T^{2} \] C1 \[ \begin{aligned} a =\left(4 \pi^{2} \times 6.4 \times 10^{6}\right) /(24 \times 60 \times 60)^{2} =0.034 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 1(c)(ii)

1(c)(ii)

Describe how the two forces acting on the object give rise to this centripetal acceleration. You may draw a diagram if you wish.

Mediumstructured2 marks

Answer

identification of the two forces acting on the object as gravitational force and (normal) contact force M1 gravitational force and normal contact force are in opposite directions, and their resultant causes the (centripetal) acceleration A1

Question 1

1

2 marks

Question 1(a)

1(a)

Explain how the force(s) on a satellite can result in the satellite being in a circular orbit around a planet.

Mediumstructured2 marks

Answer

gravitational force (of attraction between satellite and planet) B1 causes centripetal acceleration (of satellite about the planet) B1

Question 1

1

5 marks

Question 1(b)

1(b)

Two cars are moving around a horizontal circular track. One car follows path X and the other follows path Y , as shown in Fig. 1.1. The radius of path X is 318 m . Path Y is parallel to, and 27 m outside, path X . Both cars have mass 790 kg . The maximum lateral (sideways) friction force F that the cars can experience without sliding is the same for both cars.

structured5 marks

Question 1(b)(i)

1(b)(i)

The maximum speed at which the car on path X can move around the track without sliding is \(94 \mathrm{~m} \mathrm{~s}^{-1}\). Calculate F. F= N

Mediumstructured2 marks

Answer

\[ F=m v^{2} / r \] or \[ v=r \omega \text { and } F=m r \omega^{2} \] C1 \[ \begin{aligned} F =790 \times 94^{2} / 318 =22000 \mathrm{~N} \end{aligned} \] A1

Question 1(b)(ii)

1(b)(ii)

Both cars move around the track. Each car has the maximum speed at which it can move without sliding. Complete Table 1.1, by placing one tick in each row, to indicate how the quantities indicated for the car on path Y compare with the car on path X .

Mediumstructured3 marks

Answer

centripetal acceleration: same B1 maximum speed: greater B1 time taken for one lap of the track: greater B1

Question 1

1

8 marks

Question 1(a)

1(a)

State what is meant by centripetal acceleration.

Easystructured1 marks

Answer

acceleration perpendicular to velocity B1

Question 1(b)

1(b)

An unpowered toy car moves freely along a smooth track that is initially horizontal. The track contains a vertical circular loop around which the car travels, as shown in Fig. 1.1. The mass of the car is 230 g and the diameter of the loop is 62 cm . Assume that the resistive forces acting on the car are negligible.

structured3 marks

Question 1(b)(i)

1(b)(i)

State what happens to the magnitude of the centripetal acceleration of the car as it moves around the loop from X to Y .

Mediumstructured1 marks

Answer

decreases B1

Question 1(b)(ii)

1(b)(ii)

Explain, if the car remains in contact with the track, why the centripetal acceleration of the car at point Y must be greater than \(9.8 \mathrm{~ms}^{-2}\).

Mediumstructured2 marks

Answer

(acceleration of) \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\) is caused by weight of car or centripetal force must be greater than weight of car B1 (acceleration \(>9.8 \mathrm{~m} \mathrm{~s}^{-2}\) ) requires contact force from track or (centripetal force > weight) requires contact force from track B1

Question 1(c)

1(c)

The initial speed at which the car in (b) moves along the track is \(3.8 \mathrm{~ms}^{-1}\). Determine whether the car is in contact with the track at point Y . Show your working.

Hardstructured3 marks

Answer

\(\frac{1}{2} m v_{\gamma^{2}}{ }^{2}=1 / 2 m v_{\mathrm{X}}{ }^{2}-m g h\) C1 \(a=v^{2} / r\) C1 \[ v_{Y}{ }^{2}=3.8^{2}-2 \times 9.81 \times 0.62 \text { so } v_{Y}=1.5 \mathrm{~ms}^{-1} \] \(a=1.5^{2} / 0.31=7.3 \mathrm{~m} \mathrm{~s}^{-2}\) (which is less than \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\) ) so no A1 or \[ v_{Y}=\sqrt{ }(9.81 \times 0.31)=1.74 \mathrm{~m} \mathrm{~s}^{-1} \text { so } v_{X}{ }^{2}=1.74^{2}+2 \times 9.81 \times 0.62 \] \(v_{\mathrm{x}}=3.9 \mathrm{~m} \mathrm{~s}^{-1}\) (which is greater than \(3.8 \mathrm{~m} \mathrm{~s}^{-1}\) ) so no (A1)

Question 1(d)

1(d)

Suggest, with a reason but without calculation, whether your conclusion in (c) would be different for a car of mass 460 g moving with the same initial speed.

Mediumstructured1 marks

Answer

acceleration is independent of mass so makes no difference or mass cancels in the equation so makes no difference B1

Question 1

1

A planet of mass m is in a circular orbit of radius r about the Sun of mass M, as illustrated in Fig. 1.1. The magnitude of the angular velocity and the period of revolution of the planet about the Sun are \(\omega\) and T respectively.

structured4 marks

Question 1(b)

1(b)

Show that, for a planet in a circular orbit of radius r, the period T of the orbit is given by the expression where c is a constant. Explain your working.

Mediumstructured4 marks

Answer

centripetal force is provided by the gravitational force B1 either \(m r(2 \pi / T)^{2}=G M m / r^{2}\) or \(m r \omega^{2}=G M m / r^{2} \quad\) M1 \(r^{3} \times 4 \pi^{2}=G M \times T^{2} \quad\) A1 \(G M / 4 \pi^{2}\) is a constant (c) A1 \(T^{2}=c r^{3}\) A0

Question 2

2

A large bowl is made from part of a hollow sphere. A small spherical ball is placed inside the bowl and is given a horizontal speed. The ball follows a horizontal circular path of constant radius, as shown in Fig. 2.1. The forces acting on the ball are its weight W and the normal reaction force R of the bowl on the ball, as shown in Fig. 2.2. The normal reaction force R is at an angle \(\theta\) to the horizontal.

structured4 marks

Question 2(a)

2(a)

1 marks

Question 2(a)(ii)

2(a)(ii)

State the significance of the force F for the motion of the ball in the bowl.

Mediumstructured1 marks

Answer

provides the centripetal force

Question 2(b)

2(b)

The ball moves in a circular path of radius 14 cm . For this radius, the angle \(\theta\) is \(28^{\circ}\). Calculate the speed of the ball. speed = \(\mathrm{ms}^{-1}\)

Mediumstructured3 marks

Answer

either \(F=m v^{2} / r\) and W=m g or \(v^{2}=r g / \tan \theta\) \(v^{2}=\left(14 \times 10^{-2} \times 9.8\right) / \tan 28^{\circ}\) =2.58 \(v=1.6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 2

2

structured4 marks

Question 2(a)

2(a)

1 marks

Question 2(a)(ii)

2(a)(ii)

State the significance of the force F for the motion of the ball in the bowl.

Mediumstructured1 marks

Answer

provides the centripetal force

Question 2(b)

2(b)

The ball moves in a circular path of radius 14 cm . For this radius, the angle \(\theta\) is \(28^{\circ}\). Calculate the speed of the ball. speed = \(\mathrm{ms}^{-1}\)

Mediumstructured3 marks

Answer

either \(F=m v^{2} / r\) and W=m g or \(v^{2}=r g / \tan \theta\) \(v^{2}=\left(14 \times 10^{-2} \times 9.8\right) / \tan 28^{\circ}\) =2.58 \(v=1.6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 2

2

A steel sphere of mass 0.29 kg is suspended in equilibrium from a vertical spring. The centre of the sphere is 8.5 cm from the top of the spring, as shown in Fig. 2.1. The sphere is now set in motion so that it is moving in a horizontal circle at constant speed, as shown in Fig. 2.2. The distance from the centre of the sphere to the top of the spring is now 10.8 cm .

structured7 marks

Question 2(a)

2(a)

Explain, with reference to the forces acting on the sphere, why the length of the spring in Fig. 2.2 is greater than in Fig. 2.1.

Mediumstructured3 marks

Answer

horizontal force on sphere causes centripetal acceleration B1 weight of sphere is (now) equal to vertical component of tension or horizontal and vertical components (of force) (now) combine to give greater tension (in spring) B1 greater tension in spring so greater extension of spring B1

Question 2(b)

2(b)

The angle between the linear axis of the spring and the vertical is \(27^{\circ}\).

structured2 marks

Question 2(b)(ii)

2(b)(ii)

Show that the tension in the spring is 3.2 N .

Mediumstructured2 marks

Answer

\[ T \cos \theta=m g \] or \[ T \cos \theta=W \text { and } W=m g \] C1 \(T \cos 27^{\circ}=0.29 \times 9.81\) leading to \(T=3.2 \mathrm{~N}\) A1

Question 2(c)

2(c)

2 marks

Question 2(c)(i)

2(c)(i)

Use the information in (b) to determine the centripetal acceleration of the sphere.

Mediumstructured2 marks

Answer

centripetal acceleration \(=(T \sin \theta) / m\) \[ =\left(3.2 \times \sin 27^{\circ}\right) / 0.29 \] C1 \(=5.0 \mathrm{~m} \mathrm{~s}^{-2}\) A1