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A-Level CAIE Physics 2 1 Equations Of Motion Question Bank

Practice A-Level CAIE Physics 2 1 Equations 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

8 marks

Question 1(b)

1(b)

Under certain conditions, the distance s moved in a straight line by an object in time t is given by where a is the acceleration of the object. State two conditions under which the above expression applies to the motion of the object. 1 2

Mediumstructured2 marks

Answer

initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1

Question 1(c)

1(c)

The variation with time t of the velocity v of a car that is moving in a straight line is shown in Fig. 1.1.

structured6 marks

Question 1(c)(i)

1(c)(i)

Compare, qualitatively, the acceleration of the car at time \(t=8.0 \mathrm{~s}\) and at time \(t=14.0 \mathrm{~s}\) in terms of: - magnitude - direction.

Mediumstructured2 marks

Answer

magnitude of acceleration at \(t=8.0 \mathrm{~s}\) is less than that at \(t=14.0 \mathrm{~s}\) B1 - direction of acceleration at \(t=8.0 \mathrm{~s}\) is opposite to that at \(t=14.0 \mathrm{~s}\) B1

Question 1(c)(ii)

1(c)(ii)

Determine the magnitude of the acceleration of the car at time \(t=4.0 \mathrm{~s}\). acceleration = \(\mathrm{ms}^{-2}\)

Mediumstructured2 marks

Answer

a= gradient or a=(v-u) / t or \(a=\Delta v /(\Delta) t\) C1 \[ \begin{aligned} a=\text { e.g. }(20+10) / 12 \text { or }(0+10) / 4 \text { or }(20-0) /(12-4) a=2.5 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 1(c)(iii)

1(c)(iii)

The car is at point X at time t=0. Determine the magnitude of the displacement of the car from X at time \(t=12.0 \mathrm{~s}\). displacement = m

Mediumstructured2 marks

Answer

\[ \text { displacement }=[1 / 2 \times(12-4) \times 20]-[1 / 2 \times 4 \times 10] \] or displacement \(=(-10 \times 12)+\left(1 / 2 \times 2.5 \times 12^{2}\right)\) or displacement \(=(20 \times 12)-\left(\frac{1}{2} \times 2.5 \times 12^{2}\right)\) or displacement \(=1 / 2 \times(20-10) \times 12\) C1 displacement \(=60 \mathrm{~m}\) A1

Question 1

1

A well has a depth of 36 m from ground level to the surface of the water in the well, as shown in Fig. 1.1. A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water.

structured7 marks

Question 1(a)

1(a)

Assume that air resistance is negligible and that the stone is released from rest. Calculate the time taken for the stone to fall from ground level to the surface of the water. time = s

Easystructured2 marks

Answer

\[ \begin{aligned} t =\sqrt{ }(2 s / g) =\sqrt{ }[(2 \times 36) / 9.81] \end{aligned} \] C1 \(=2.7 \mathrm{~s}\) A1

Question 1

1

3 marks

Question 1(b)

1(b)

A toy car moves in a horizontal straight line. The displacement s of the car is given by the equation where a is the acceleration of the car and v is its final velocity. State two conditions that apply to the motion of the car in order for the above equation to be valid. 1 2

Mediumstructured2 marks

Answer

initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1

Question 1(c)

1(c)

An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained:

structured1 marks

Question 1(c)(i)

1(c)(i)

Calculate the acceleration a of the car.

Mediumstructured1 marks

Answer

\[ \begin{aligned} a =2.75^{2} /(2 \times 3.89) =0.97 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 1

1

2 marks

Question 1(c)

1(c)

The maximum useful output power P of a car travelling on a horizontal road is given by where v is the maximum speed of the car and b is a constant. For the car, and \(b=0.56 \pm 7 \%\) in SI units.

structured2 marks

Question 1(c)(i)

1(c)(i)

Calculate the value of v.

Mediumstructured2 marks

Answer

\(84 \times 10^{3}=v^{3} \times 0.56\) C1 \(v=53 \mathrm{~m} \mathrm{~s}^{-1}\) A1

Question 1

1

4 marks

Question 1(a)

1(a)

Define velocity.

Easystructured1 marks

Answer

change in displacement / time (taken) B1

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.

structured3 marks

Question 1(b)(i)

1(b)(i)

Determine the height of the rock above the surface of the planet at time \(t=4.0 \mathrm{~s}\). height = m

Mediumstructured3 marks

Answer

(displacement =) area under graph C1 (at \(t=4.0 \mathrm{~s}\) ) v=(-) 2.4 C1 \[ \begin{aligned} \text { height } =1 / 2 \times 2.5 \times 4.0-1 / 2 \times 1.5 \times 2.4 =3.2 \mathrm{~m} \end{aligned} \] A1

Question 1

1

1 marks

Question 1(a)

1(a)

1 marks

Question 1(a)(i)

1(a)(i)

Define velocity.

Easystructured1 marks

Answer

either rate of change of displacement or (change in) displacement/time (taken) B1

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,

structured0 marks

Question 1(b)(i)

1(b)(i)

calculate the distance moved by the car, m

Mediumstructured0 marks

Answer

idea of area under graph/use of \(s=\frac{(u+v)}{2} \times t \quad\) C1

Question 1

1

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

structured2 marks

Question 1(b)

1(b)

A raindrop falls vertically from rest. Assume that air resistance is negligible.

structured3 marks

Question 1(b)(i)

1(b)(i)

On Fig. 1.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion.

Mediumstructured1 marks

Answer

straight line from (0,0) to \((1,9.8) \pm\) half a square B1

Question 1(b)(ii)

1(b)(ii)

Calculate the velocity of the raindrop after falling 1000 m . velocity = \(\mathrm{ms}^{-1}\)

Hardstructured2 marks

Answer

\(\frac{1}{2} m v^{2}=m g h \quad\) or using \(v^{2}=2\) as C1 \(v=(2 \times 9.81 \times 1000)^{1 / 2}=140 \mathrm{~m} \mathrm{~s}^{-1} \quad\) A1 [2]

Question 1

1

3 marks

Question 1(a)

1(a)

1 marks

Question 1(a)(i)

1(a)(i)

Define acceleration.

Easystructured1 marks

Answer

acceleration = change in velocity / time (taken) or acceleration = rate of change of velocity 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.

structured2 marks

Question 1(b)(i)

1(b)(i)

Calculate the distance travelled by the parachutist in the first 3.0 s of the motion. distance = m

Mediumstructured2 marks

Answer

distance is represented by the area under graph C1 distance \(=1 / 2 \times 29.5 \times 3=44.3 \mathrm{~m}\) (accept 43.5 m for 29 to 45 m for 30) A1 [2]

Question 1

1

The variation with time t of the displacement s for a car is shown in Fig. 1.1.

structured6 marks

Question 1(a)

1(a)

Determine the magnitude of the average velocity between the times 5.0 s and 35.0 s . average velocity = \(\mathrm{ms}^{-1}\)

Easystructured2 marks

Answer

average velocity =540 / 30 \(=18 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 1(b)

1(b)

On Fig. 1.2, sketch the variation with time t of the velocity v for the car.

Mediumstructured4 marks

Answer

velocity zero at time t=0 positive value and horizontal line for time \(t=5 \mathrm{~s}\) to 35 s line / curve through v=0 at \(t=45 \mathrm{~s}\) to negative velocity negative horizontal line from 53 s with magnitude less than positive value and horizontal line to time \(=100 \mathrm{~s}\)

Question 1

1

the magnitude of the frictional force F,

structured7 marks

Question 1(c)

1(c)

A stone is thrown with a horizontal velocity of \(20 \mathrm{~ms}^{-1}\) from the top of a cliff 15 m high. The path of the stone is shown in Fig. 1.1. Air resistance is negligible. For this stone,

structured7 marks

Question 1(c)(i)

1(c)(i)

calculate the time to fall 15 m , time = s

Mediumstructured2 marks

Answer

\(s=u t+1 / 2 a t^{2}\) \(15=0.5 \times 9.81 \times t^{2} \quad\) C1 \(T=1.7 \mathrm{~s}\) A1 if g=10 is used then -1 but only once on paper

Question 1(c)(ii)

1(c)(ii)

calculate the magnitude of the resultant velocity after falling 15 m , \(\mathrm{ms}^{-1}[3]\)

Mediumstructured3 marks

Answer

vertical component \(v_{\mathrm{v}}\) : \(v_{\mathrm{v}}^{2}=u^{2}+2 a s=0+2 \times 9.81 \times 15\) or \(v_{\mathrm{v}}=u+a t=9.81 \times 1.7(5)\) \(v_{\mathrm{v}}=17.16\) resultant velocity: \(v^{2}=(17.16)^{2}+(20)^{2}\) \(v=26 \mathrm{~m} \mathrm{~s}^{-1}\) If u=20 is used instead of u=0 then 0 / 3 Allow the solution using: initial (potential energy + kinetic energy) = final kinetic energy

Question 1(c)(iii)

1(c)(iii)

describe the difference between the displacement of the stone and the distance that it travels.

Mediumstructured2 marks

Answer

distance is the actual path travelled B1 displacement is the straight line distance between start and finish points (in that direction) / minimum distance B1