ENGAA 2017

• A.1
• B.3
• C.$\frac{5}{3}$
• D.$\frac{7}{3}$
• E.$3\sqrt{3}$
• F.9

Answer: F

A car is travelling along a horizontal road in a straight line.

The graph is a velocity-time graph for part of the car's journey.

(Image of a velocity-time graph is shown. Key points: (0,20), (30,20), (50,30), (110,30), (120,20), (150,20))

During this part of the journey, what is the total distance that the car travels while it is decelerating?

• A.400 m
• B.500 m
• C.550 m
• D.600 m
• E.750 m
• F.1400 m
• G.1800 m
• H.1900 m

Answer: B

• A.$x \leq -3$
• B.$x \geq 2.5$
• C.$x \leq -1.5, x \geq 5$
• D.$-1.5 \leq x \leq 5$
• E.$x \leq -3, x \geq 2.5$
• F.$-3 \leq x \leq 2.5$

Answer: E

• A.none of them
• B.1 only
• C.2 only
• D.3 only
• E.1 and 2 only
• F.1 and 3 only
• G.2 and 3 only
• H.1, 2 and 3

Answer: G

• A.$x=2\pm2\sqrt{\frac{y-5}{3}}$
• B.$x=2\pm2\sqrt{\frac{y+5}{3}}$
• C.$x=2\pm3\sqrt{\frac{y+5}{3}}$
• D.$x=-2\pm2\sqrt{\frac{y+5}{3}}$
• E.$x=-2\pm3\sqrt{\frac{y+5}{2}}$
• F.$x=2+2\left(\frac{y+5}{3}\right)^2$
• G.$x=-2+2\left(\frac{y+5}{3}\right)^2$

Answer: B

An electric motor is used to pull a broken-down car slowly from the road up a ramp on to the back of a breakdown truck.

The car has a mass of 1200 kg and is lifted through a vertical height of 1.0 m.

The total input energy to the motor is 28 kJ and it is 75% efficient.

In the process of lifting the car, energy is lost to the surroundings from the motor and from other causes.

What is the total energy lost to the surroundings?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.7.0 kJ
• B.9.0 kJ
• C.12 kJ
• D.16 kJ
• E.21 kJ
• F.33 kJ

Answer: D

• A.$\frac{2Q-3P}{3}$
• B.$\frac{2Q-3P}{11}$
• C.$\frac{Q-P}{3}$
• D.$\frac{Q-P}{11}$
• E.$\frac{P-Q}{3}$
• F.$\frac{3P-2Q}{3}$
• G.$\frac{3P-2Q}{11}$

Answer: G

• A.6.0 mm
• B.12 mm
• C.24 mm
• D.6.0 cm
• E.12 cm
• F.24 cm

Answer: D

• A.$P = \frac{5}{R}$
• B.$P = \frac{5}{4R}$
• C.$P = \frac{1}{800R^2}$
• D.$P = \frac{5}{4R^2}$
• E.$P = \frac{25}{2R^2}$
• F.$P = \frac{800}{R^2}$
• G.$P = \frac{R^2}{50}$
• H.$P = \frac{25R^2}{2}$

Answer: E

• A.$w + y = 240$
• B.$z = 240 - (w + y)$
• C.$x = 40 - z$
• D.$94 = 54 + x + 1$
• E.$240 = 54 + x$
• F.$94 = w + y + 1$

Answer: B

• A.$-1 - \frac{2}{x}$
• B.$-1 + \frac{2}{x}$
• C.$5 - \frac{2}{x}$
• D.$5 + \frac{2}{x}$
• E.$5 - \frac{3}{x}$
• F.$5 + \frac{3}{x}$

Answer: D

An electric motor is connected to a constant 12V d.c. supply. The motor is used to lift a mass of 20 kg by means of a rope and pulley. The mass is lifted vertically through a height of 6.0 m in a time of 5.0 s. The complete lifting system (motor, rope and pulley) is 80% efficient.

What is the current in the electric motor?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.1.6 A
• B.2.0 A
• C.2.5 A
• D.16 A
• E.20 A
• F.25 A

Answer: F

• A.-2.25
• B.-1.75
• C.-1.5
• D.-0.5
• E.-0.25

Answer: D

The nuclide $_Q^P X$ decays to the stable nuclide Y. During this process four particles are emitted: an $\alpha$-particle and three $\beta^-$ particles.

Which of the following is not a nuclide that could be formed at any stage during this process?

• A.nuclide with atomic mass P, atomic number Q-1
• B.nuclide with atomic mass P, atomic number Q+1
• C.nuclide with atomic mass P, atomic number Q+2
• D.nuclide with atomic mass P, atomic number Q+3
• E.nuclide with atomic mass P-4, atomic number Q-2
• F.nuclide with atomic mass P-4, atomic number Q-1
• G.nuclide with atomic mass P-4, atomic number Q
• H.nuclide with atomic mass P-4, atomic number Q+1

Answer: A

• A.X+2Y
• B.X+Y+35
• C.X+3Y-35
• D.2X+2Y
• E.2X + Y - 35
• F.2X+3Y-35
• G.2X + Y + 35

Answer: F

• A.$(3.0 \times 10^4)^{-3}$
• B.$(3.0 \times 10^4)^{-2}$
• C.$(3.0 \times 10^4)^{-1}$
• D.$(3.0 \times 10^4)^1$
• E.$(3.0 \times 10^4)^2$
• F.$(3.0 \times 10^4)^3$

Answer: A

• A.10
• B.12
• C.15
• D.18
• E.21
• F.24
• G.27

Answer: C

Graph 1 shows how the displacement of one of the particles of a medium varies with time in seconds as a wave travels through the medium.

Graph 2 shows how the displacement varies with distance in metres at one time for the same wave.

Which expression gives the speed in $\text{ms}^{-1}$ of the wave?

• A.$\frac{4(x_2-x_1)}{3(t_2-t_1)}$
• B.$\frac{3(x_2-x_1)}{2(t_2-t_1)}$
• C.$\frac{2(x_2-x_1)}{t_2-t_1}$
• D.$\frac{8(x_2-x_1)}{3(t_2-t_1)}$
• E.$\frac{3(x_2-x_1)}{t_2-t_1}$
• F.$\frac{6(x_2-x_1)}{t_2-t_1}$

Answer: E

• A.$070^\circ$
• B.$080^\circ$
• C.$090^\circ$
• D.$100^\circ$
• E.$140^\circ$

Answer: B

• A.96 C
• B.250 C
• C.15000 C
• D.36000 C
• E.900000 C
• F.2160000 C

Answer: C

• A.$80^\circ$
• B.$100^\circ$
• C.$110^\circ$
• D.$120^\circ$
• E.$140^\circ$

Answer: B

• A.7
• B.9
• C.10
• D.24
• E.46

Answer: C

• A.2
• B.4
• C.6
• D.8
• E.9
• F.11
• G.12

Answer: C

• A.none of them
• B.1 only
• C.2 only
• D.3 only
• E.1 and 2 only
• F.1 and 3 only
• G.2 and 3 only
• H.1, 2 and 3

Answer: C

The diagram shows a square with side of length $x$ cm. A circle is drawn with centre O which lies at the mid-point of one of the sides of the square. This side forms part of a diameter of the circle. The circle passes through two corners of the square as shown.

What is the area, in cm$^2$, of the shaded part of the semi-circle?

• A.$(\pi-1)x^2$
• B.$(\frac{\pi-2}{2})x^2$
• C.$(\frac{3\pi-2}{2})x^2$
• D.$(\frac{3\pi-4}{4})x^2$
• E.$(\frac{5\pi-4}{4})x^2$
• F.$(\frac{5\pi-8}{8})x^2$

Answer: F

• A.$\frac{10}{16}$
• B.$\frac{11}{16}$
• C.$\frac{12}{16}$
• D.$\frac{13}{16}$
• E.$\frac{14}{16}$
• F.$\frac{15}{16}$

Answer: D

A cylindrical hollow metal pipe is 16 cm long.

It has an external diameter of 10 cm and an internal diameter of 8 cm.

The density of the metal from which the pipe is made is 8 grams per cm$^3$.

[diagram not to scale]

What is the mass of the pipe in grams?

• A.$8\pi$
• B.$16\pi$
• C.$18\pi$
• D.$72\pi$
• E.$128\pi$
• F.$512\pi$
• G.$1152\pi$
• H.$4608\pi$

Answer: G

• A.kinetic energy of car X = $0.90 \times$ kinetic energy of car Y
• B.kinetic energy of car X = $0.96 \times$ kinetic energy of car Y
• C.kinetic energy of car X = $1.20 \times$ kinetic energy of car Y
• D.kinetic energy of car X = $1.44 \times$ kinetic energy of car Y
• E.kinetic energy of car X = $1.80 \times$ kinetic energy of car Y

Answer: E

• A.$-\frac{3}{4}$
• B.$\frac{3}{4}$
• C.$-\frac{3}{4} - \frac{\sqrt{3}}{7}$
• D.$\frac{3}{4} - \frac{\sqrt{3}}{7}$
• E.$-\frac{3}{4} - \sqrt{3}$
• F.$\frac{3}{4} - \sqrt{3}$
• G.$-\frac{\sqrt{3}}{2}$
• H.$\frac{\sqrt{3}}{2}$

Answer: E

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

The diagram shows a crane being used on a building site. The crane is perfectly balanced about P.

[diagram not to scale, shows a load of 400kg at 10m left of P and a counterweight of 2000kg at 10m right of P. The weight of the horizontal section of the crane is also a factor.]

The load is now moved to the left by 5.0 m.

To keep the crane perfectly balanced about P, how far does the counterweight have to move, and in which direction?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.1.0 m to the left
• B.1.0 m to the right
• C.3.0 m to the left
• D.3.0 m to the right
• E.4.0 m to the left
• F.4.0 m to the right

Answer: B

• A.0
• B.2
• C.3
• D.4
• E.8

Answer: C

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

A ball is thrown vertically upwards in still air and is then caught at the same height when it comes back down.

Which velocity-time graph shows this complete motion?

(Take upwards as positive, and ignore air resistance.)

• A.Graph of velocity vs time: straight line with negative slope, starting with positive velocity, crossing time axis, ending with negative velocity.
• B.Graph of velocity vs time: downward-opening parabola starting with positive velocity.
• C.Graph of velocity vs time: sine wave starting at a positive peak.
• D.Graph of velocity vs time: straight line with negative slope, starting with positive velocity, ending at zero velocity.
• E.Graph of velocity vs time: 'V' shape, negative slope to zero velocity, then positive slope.
• F.Graph of velocity vs time: step function, positive constant velocity then negative constant velocity.
• G.Graph of velocity vs time: 'V' shape, positive slope from negative velocity to zero, then positive slope to positive velocity.
• H.Graph of velocity vs time: sine wave starting from origin.

Answer: A

• A.$\log_2 3$
• B.$\log_3 2$
• C.2
• D.$\log_{10} 2$
• E.$\frac{2}{3}$

Answer: B

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

• A.It is parallel to its motion.
• B.It is perpendicular to its motion.
• C.It is zero.
• D.It is equal to its weight.
• E.It is equal to the drag acting on the aircraft.

Answer: C

The diagram shows the outline of a keyhole consisting of three straight sides and an arc from a circle.

The sides PQ and RS are both 18 mm in length and when extended meet at the centre of the circle O forming an angle of $\frac{\pi}{6}$ radians.

The longer arc from Q to R has length $22\pi$ mm.

[diagram not to scale]

What is the area, in mm$^2$, of the keyhole as shaded in the diagram?

• A.$121\pi + \frac{841}{4}$
• B.$121\pi + \frac{841\sqrt{3}}{4}$
• C.$132\pi + 225$
• D.$132\pi + 225\sqrt{3}$
• E.$144\pi + 225$
• F.$144\pi + 225\sqrt{3}$

Answer: C

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

A horizontal, uniform bar of mass 60 kg is 4.0 m long and is pivoted at one end. The bar is held in equilibrium by a force $F$ at the other end of the bar, acting at an angle of $60^\circ$ to the horizontal.

[diagram not to scale]

Which expression gives the magnitude of $F$ in newtons?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.$\frac{30}{\sin 60^\circ}$
• B.$\frac{30}{\cos 60^\circ}$
• C.$\frac{60}{\sin 60^\circ}$
• D.$\frac{60}{\cos 60^\circ}$
• E.$\frac{300}{\sin 60^\circ}$
• F.$\frac{300}{\cos 60^\circ}$
• G.$\frac{600}{\sin 60^\circ}$
• H.$\frac{600}{\cos 60^\circ}$

Answer: E

• A.$6p-4q$
• B.$6p+4q$
• C.$6p-8q$
• D.$6p+8q$
• E.$9p-4q$
• F.$9p+4q$
• G.$9p-8q$
• H.$9p+8q$

Answer: F

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

• A.$16.0 \text{ms}^{-1}$
• B.$20.0 \text{ms}^{-1}$
• C.$25.6 \text{ms}^{-1}$
• D.$32.0 \text{ms}^{-1}$
• E.$36.2 \text{ms}^{-1}$
• F.$39.3 \text{ms}^{-1}$

Answer: D

• A.$p=-3$ and $q=-8$
• B.$p=-3$ and $q=8$
• C.$p=3$ and $q=-8$
• D.$p=-9$ and $q=24$
• E.$p=9$ and $q=24$
• F.$p=9$ and $q=-24$

Answer: D

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

A block of mass 1.0 kg is at rest on a rough horizontal surface. The block is attached by a light inextensible string to a force meter. The other end of the force meter is attached by another light inextensible string via a frictionless pulley to a load of mass 1.0 kg. The block remains stationary.

What is the reading on the force meter?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.0.0 N
• B.0.5 N
• C.1.0 N
• D.2.0 N
• E.5.0 N
• F.10 N
• G.20 N

Answer: F

• A.$\frac{8\sqrt{3}}{3}$
• B.$\frac{16}{3}$
• C.$\frac{69\sqrt{3}}{16}$
• D.$\frac{16\sqrt{3}}{3}$
• E.$\frac{32}{3}$
• F.$\frac{32\sqrt{3}}{3}$

Answer: D

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

• A.0.80 J
• B.3.6 J
• C.4.0 J
• D.7.2 J
• E.8.0 J

Answer: A

• A.240
• B.4320
• C.4860
• D.12960
• E.19440

Answer: E

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

A stone is fired vertically upwards at a speed of $13 \text{ms}^{-1}$ on a still day from the top of a 6.0 m high cliff. It then falls down and lands at the bottom of the cliff.

[diagram not to scale]

From when the stone passes the top of the cliff on the way down, how long does it take to reach the ground at the bottom of the cliff?

(air resistance can be ignored; gravitational field strength = $10 \text{N kg}^{-1}$)

• A.0.40 s
• B.$\frac{6.0}{6.5}$ s
• C.0.60 s
• D.$\sqrt{1.2}$ s
• E.1.3 s
• F.2.0 s
• G.2.5 s
• H.3.0 s

Answer: A

• A.$\frac{13}{12}\pi, \frac{17}{12}\pi$
• B.$\frac{7}{6}\pi, \frac{4}{3}\pi$
• C.$\frac{7}{6}\pi, \frac{11}{6}\pi$
• D.$\frac{5}{4}\pi, \frac{7}{4}\pi$
• E.there are no values of x in this range

Answer: A

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

An archer fires an arrow of mass 0.024 kg vertically upwards from a bow.

The graph shows how the force of the bowstring on the arrow varies with distance as the arrow moves upwards.

(Graph is linear, from (0, 192) to (0.40, 0))

The work done by the force of the bowstring is given by the area under the force-distance graph.

When the arrow leaves the bow, what is the kinetic energy of the arrow, and what is the maximum height that it gains from this point?

(Air resistance can be ignored. The effect of gravity as the arrow is fired is negligible compared to the force of the bowstring. The gravitational field strength = $10 \text{N kg}^{-1}$.)

• A.kinetic energy / J: 38.4, height / m: 16
• B.kinetic energy / J: 38.4, height / m: 160
• C.kinetic energy / J: 38.4, height / m: 1600
• D.kinetic energy / J: 38.4, height / m: 16000
• E.kinetic energy / J: 76.8, height / m: 32
• F.kinetic energy / J: 76.8, height / m: 320
• G.kinetic energy / J: 76.8, height / m: 3200
• H.kinetic energy / J: 76.8, height / m: 32000

Answer: B

• A.-10
• B.-2
• C.-1
• D.8
• E.26

Answer: C

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

A book of mass $m$ rests on a rough horizontal surface. The surface is now tilted as shown:

When the angle of tilt $\theta$ is $20^\circ$, the book slides down the slope at a constant speed.

What is the acceleration of the book down the slope when the angle of tilt is $25^\circ$?

(gravitational field strength = $g$)

• A.$g(\cos 20^\circ - \sin 20^\circ \tan 5^\circ)$
• B.$g(\cos 20^\circ - \sin 20^\circ \tan 25^\circ)$
• C.$g(\cos 25^\circ - \sin 5^\circ \tan 20^\circ)$
• D.$g(\cos 25^\circ - \sin 25^\circ \tan 20^\circ)$
• E.$g(\sin 20^\circ - \cos 20^\circ \tan 5^\circ)$
• F.$g(\sin 20^\circ - \cos 20^\circ \tan 25^\circ)$
• G.$g(\sin 25^\circ - \cos 5^\circ \tan 20^\circ)$
• H.$g(\sin 25^\circ - \cos 25^\circ \tan 20^\circ)$

Answer: H

• A.$x<0, x>4$
• B.$0<x<4$
• C.$0<x<1, x>4$
• D.$x<0, 1<x<4$
• E.$x<1, x>4$
• F.$1<x<4$

Answer: A

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

• A.$\mu mg$
• B.$mg \sin \theta$
• C.$mg \cos \theta$
• D.$\mu mg \sin \theta$
• E.$\mu mg \cos \theta$

Answer: B

• A.$y=\sin(\frac{x}{2}-\frac{\pi}{4})$
• B.$y=\sin(\frac{x}{2}+\frac{\pi}{4})$
• C.$y=\sin(\frac{x}{2}-\frac{\pi}{8})$
• D.$y=\sin(\frac{x}{2}+\frac{\pi}{8})$
• E.$y=\sin(2x-\frac{\pi}{4})$
• F.$y=\sin(2x+\frac{\pi}{4})$
• G.$y=\sin(2x-\frac{\pi}{2})$
• H.$y=\sin(2x+\frac{\pi}{2})$

Answer: H

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

The graph shows how the horizontal force on a tennis ball of mass $m$ varies during a shot in a tennis match. The ball is initially travelling horizontally toward the racket with speed $u$ and leaves the racket horizontally travelling in the opposite direction with speed $v$.

Which expression gives the magnitude of the momentum of the ball as it leaves the racket?

• A.$F(t_2-t_1)$
• B.$F(t_2-t_1) - mu$
• C.$F(t_2-t_1) + mu$
• D.$mv - mu$
• E.$Ft_2 - mu$

Answer: B

• A.2.00
• B.1.75
• C.1.50
• D.1.00
• E.-0.25
• F.-0.50

Answer: B

Not in the current ESAT specification — included for completeness, but you won't be tested on it.

The acceleration versus time graph is for a ball dropped from rest, falling vertically and bouncing on the ground.

(Graph shows acceleration at -10 m/s^2, with vertical spikes at t=0.5s and t=1.3s indicating bounces. The time of contact with the ground can be ignored.)

What is the speed of the ball immediately after hitting the ground for the first time, and what is the maximum height reached by the ball after the first bounce?

(gravitational field strength = $10 \text{N kg}^{-1}$)

• A.speed / $\text{ms}^{-1}$: 4.00, height / m: 0.80
• B.speed / $\text{ms}^{-1}$: 4.00, height / m: 1.25
• C.speed / $\text{ms}^{-1}$: 5.00, height / m: 0.80
• D.speed / $\text{ms}^{-1}$: 5.00, height / m: 1.25
• E.speed / $\text{ms}^{-1}$: 8.00, height / m: 3.20

Answer: A