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A-Level CAIE Physics 3 1 Momentum And Newton S Laws Of Motion Question Bank

Practice A-Level CAIE Physics 3 1 Momentum And Newton S Laws Of Motion questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

10 matching questions · Open interactive library

Question 1

1

The drag force \(F_{\mathrm{D}}\) acting on an object falling through air is given by where A is the cross-sectional area of the object, v is the velocity of the object in the air, \(\rho\) is the density of the air and C is a constant called the drag coefficient.

structured10 marks

Question 1(c)

1(c)

The mass of the sphere is 49 g . Calculate the drag force \(F_{\mathrm{D}}\) acting on the sphere.

Mediumstructured2 marks

Answer

(at terminal velocity) \(F_{\mathrm{D}}=m g\) C1 \[ \begin{aligned} F_{D} =0.049 \times 9.81 =0.48 \mathrm{~N} \end{aligned} \] A1

Question 1

1

4 marks

Question 1(b)

1(b)

A rock of mass 7.5 kg is projected vertically upwards from the surface of a planet. The rock leaves the surface of the planet with a speed of \(4.0 \mathrm{~ms}^{-1}\) at time t=0. The variation with time t of the velocity v of the rock is shown in Fig. 1.1. Assume that the planet does not have an atmosphere and that the viscous force acting on the rock is always zero.

structured4 marks

Question 1(b)(ii)

1(b)(ii)

Determine the change in the momentum of the rock from time t=0 to time \(t=4.0 \mathrm{~s}\). change in momentum = Ns

Mediumstructured2 marks

Answer

change in momentum =7.5(-4.0-2.4) C1 \(=(-) 48 \mathrm{~N} \mathrm{~s}\) A1

Question 1(b)(iii)

1(b)(iii)

Determine the weight W of the rock on this planet. W= N

Easystructured2 marks

Answer

\(W=\Delta p /(\Delta) t\) or \(\Delta m v /(\Delta) t\) C1 \[ \begin{aligned} =48 / 4.0 =12 \mathrm{~N} \end{aligned} \] A1 or W=m a or m g or m(v-u) / t (C1) \[ \begin{aligned} =7.5 \times 1.6 \text { or } 7.5 \times(4+2.4) / 4.0 =12 \mathrm{~N} \end{aligned} \] (A1)

Question 1

1

3 marks

Question 1(c)

1(c)

An object B is on a horizontal surface. Two forces act on B in this horizontal plane. A vector diagram for these forces is shown to scale in Fig. 1.1. A force of 7.5 N towards north and a force of 2.5 N from \(30^{\circ}\) north of east act on B. The mass of B is 750 g .

structured3 marks

Question 1(c)(ii)

1(c)(ii)

1. Show that the magnitude of the resultant force on B is 6.6 N . 2. Calculate the magnitude of the acceleration of B produced by this resultant force. magnitude = \(\mathrm{ms}^{-2}[2]\)

Mediumstructured3 marks

Answer

1. correct vector triangle or working to show magnitude of resultant force \(=6.6 \mathrm{~N}\) allow 6.5 to 6.7 N if scale diagram 2. magnitude of acceleration =6.6 / 0.75 [scale diagram: (6.5 to 6.7) / 0.75]

Question 1

1

4 marks

Question 1(b)

1(b)

A car of mass 1500 kg moves along a straight, horizontal road. The variation with time t of the velocity v for the car is shown in Fig. 1.1. The brakes of the car are applied from \(t=1.0 \mathrm{~s}\) to \(t=3.5 \mathrm{~s}\). For the time when the brakes are applied,

structured3 marks

Question 1(b)(ii)

1(b)(ii)

calculate the magnitude of the resultant force on the car.

Mediumstructured3 marks

Answer

\(a=(18-32) / 2.5(=-5.6) \quad \mathrm{C} 1\) F=m a C1 \(F=1500 \times(-) 5.6=(-) 8400 \mathrm{~N}\) A1

Question 1(c)

1(c)

The direction of motion of the car in (b) at time \(t=2.0 \mathrm{~s}\) is shown in Fig. 1.2. On Fig. 1.2, show with arrows the directions of the acceleration (label this arrow A) and the resultant force (label this arrow F).

Mediumstructured1 marks

Answer

arrow labelled A and arrow labelled F both to the left B1

Question 1

1

Show that the terminal velocity of the raindrop is about \(7 \mathrm{~ms}^{-1}\).

structured2 marks

Question 1(c)

1(c)

In practice, air resistance on raindrops is not negligible because there is a drag force. This drag force is given by the expression in (a).

structured1 marks

Question 1(c)(i)

1(c)(i)

State an equation relating the forces acting on the raindrop when it is falling at terminal velocity.

Easystructured1 marks

Answer

weight = drag (D)(+ upthrust \() \quad\) B1 [1] Allow m g or W for weight and D or expression for D for drag

Question 1(c)(ii)

1(c)(ii)

The raindrop has mass \(1.4 \times 10^{-5} \mathrm{~kg}\) and cross-sectional area \(7.1 \times 10^{-6} \mathrm{~m}^{2}\). The density of the air is \(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\) and the initial velocity of the raindrop is zero. The value of C is 0.60 .

Hardstructured0 marks

Answer

1. \(m g=1.4 \times 10^{-5} \times 9.81\) C1 \(1.4 \times 10^{-5} \times 9.81=0.5 \times 0.6 \times 1.2 \times 7.1 \times 10^{-6} \times \mathrm{v}^{2} \quad\) M1 \(v=7.33 \mathrm{~m} \mathrm{~s}^{-1} \quad\) A0 2. line from (0,0) correct curvature to a horizontal line at velocity of \(7 \mathrm{~m} \mathrm{~s}^{-1} \quad\) M1 line reaches \(7 \mathrm{~m} \mathrm{~s}^{-1}\) between 1.5 s and 3.5 s A1 [2]

Question 1

1

6 marks

Question 1(a)

1(a)

1 marks

Question 1(a)(ii)

1(a)(ii)

State Newton's first law of motion.

Easystructured1 marks

Answer

body continues at constant velocity unless acted on by a resultant force B1 [1]

Question 1(b)

1(b)

The variation with time t of vertical speed v of a parachutist falling from an aircraft is shown in Fig. 1.1.

structured5 marks

Question 1(b)(iv)

1(b)(iv)

The mass of the parachutist is 95 kg . Calculate, for the parachutist between \(t=15 \mathrm{~s}\) (point C) and \(t=17 \mathrm{~s}\) (point D), 1. the average acceleration, acceleration = \(\mathrm{ms}^{-2}\) [2] 2. the average frictional force. frictional force = N [3] Please turn over for Question 2.

Mediumstructured5 marks

Answer

1. acceleration \(=\left(v_{2}-v_{1}\right) / t=(20-50) /(17-15) \quad\) C1 \(=(-) 15 \mathrm{~m} \mathrm{~s}^{-2}\) A1 [2] 2. W-F=m a C1 \(W=95 \times 9.81(=932) \quad\) C1 \(F=(95 \times 15)+932=2400(2360)(2357) \mathrm{N} \quad\) A1 [3]

Question 1

1

1 marks

Question 1(b)

1(b)

State Newton's first law.

Easystructured1 marks

Answer

a body continues at rest or at constant velocity unless acted on by a resultant (external) force

Question 1

1

The planet Mars may be considered to be an isolated sphere of diameter \(6.79 \times 10^{6} \mathrm{~m}\) with its mass of \(6.42 \times 10^{23} \mathrm{~kg}\) concentrated at its centre. A rock of mass 1.40 kg rests on the surface of Mars. For this rock,

structured1 marks

Question 1(a)

1(a)

1 marks

Question 1(a)(i)

1(a)(i)

determine its weight, weight =

Easystructured1 marks

Answer

weight \(=G M m / r^{2}=r^{-11} \times 6.42 \times 10^{23} \times 140 /\left(1 / 2 \times 679 \times 10^{6}\right)^{2} \quad\) C1

Question 2

2

A car of mass 850 kg is travelling in a horizontal straight line. The diagram shows the two horizontal forces acting on the car in opposite directions. One force has magnitude 1200 N , and the other force has magnitude 1600 N . What is the magnitude of the acceleration of the car? \(0.47 \mathrm{~ms}^{-2}\) \(1.4 \mathrm{~m} \mathrm{~s}^{-2}\) \(1.9 \mathrm{~m} \mathrm{~s}^{-2}\) \(3.3 \mathrm{~m} \mathrm{~s}^{-2}\)

Easymcq1 marks

Answer

Question 4

4

An object of fixed mass is initially at rest at point P. The object then moves away from point P with uniform acceleration. Which statement describes the resultant force acting on the object when it is moving? It increases uniformly with respect to time. It is constant but not zero. It is proportional to the displacement from point P. It is zero.

Easymcq1 marks

Answer