CAIE IGCSE Physics 1.2 Motion Questions

[Maximum number: 3]

Fig. 1.1 shows a straight section of a river where the water is flowing from right to left at a speed of $0.54 \mathrm{~m} / \mathrm{s}$ .

Fig. 1.1 (not to scale)

A swimmer starts at point P and swims at a constant speed of $0.72 \mathrm{~m} / \mathrm{s}$ relative to the water and at right angles to the current.

After 1.5 minutes, the swimmer reaches point Q.

Calculate the distance between P and Q.
distance =

(distance $=) 81 \mathrm{~m}$ $v=s / t \mathbf{O R}(s=) v t \mathbf{O R}(s=) 0.9(0) \times 90$
(time =) $1.5 \times 60(=90)$ OR (time =) 90

[Maximum number: 3]

Fig. 2.1 shows a motorcyclist accelerating along a straight horizontal section of track.

Fig. 2.1

The motorcyclist and motorcycle have a combined mass of 240 kg .

(a)

On the straight horizontal section of the track, the motorcyclist accelerates from rest at $7.2 \mathrm{~m} / \mathrm{s}^{2}$ .

[ 2 ]

(i)

The motorcyclist reaches the end of the straight section of track in 5.3 s .

Calculate the speed of the motorcyclist at the end of the straight section.
speed =

[ 2 ]

(speed $=) 38 \mathrm{~m} / \mathrm{s}$ $a=\Delta v / \Delta t \mathbf{O R}(\Delta v=) a \Delta t \mathbf{O R}(\Delta v=) 7.2 \times 5.3$

(b)

At the end of the straight section, the track remains horizontal but bends to the right, as shown in Fig. 2.1.

When the motorcyclist reaches the bend, she travels around the bend in a circular path at a constant speed.

[ 1 ]

(i)

Describe what happens to the velocity of the motorcyclist as she travels around the bend at constant speed.

[ 1 ]

[Maximum number: 9]

Fig. 1.1 shows the speed-time graph of a person on a journey.
On the journey, he walks and then waits for a bus. He then travels by bus. He gets off the bus and waits for two minutes. He then walks again. His journey takes 74 minutes.

Fig. 1.1

(a)

For the whole journey calculate:

[ 5 ]

(i)

the distance travelled
distance =

[ 3 ]

s=vt in any form OR (s =) vt OR relates distance to area (under graph)
any one of: $5 \times 20 / 60$ OR $40 \times 20 / 60$ OR $6 \times 22 / 60$ $(\mathrm{s}=1.667+13.333+2.2=) 17 \mathrm{~km}$

(ii)

the average speed.
average speed =

[ 2 ]

average speed = candidate's (i) / time
(average speed $=17 \times 60 / 74=$ ) $14 \mathrm{~km} / \mathrm{h}$

(b)

State and explain which feature of a speed-time graph shows acceleration.

[ 2 ]

(c)

State and explain the acceleration of the person at time =40 minutes.
[Total: 9]

[ 2 ]

[Maximum number: 1]

Fig. 2.1 shows a motorcyclist accelerating along a straight horizontal section of track.

Fig. 2.1

The motorcyclist and motorcycle have a combined mass of 240 kg .

At the end of the straight section, the track remains horizontal but bends to the right, as shown in Fig. 2.1.

When the motorcyclist reaches the bend, she travels around the bend in a circular path at a constant speed.

Describe what happens to the velocity of the motorcyclist as she travels around the bend at constant speed.

[Maximum number: 2]

Fig. 1.1 shows the speed-time graph of a person on a journey.
On the journey, he walks and then waits for a bus. He then travels by bus. He gets off the bus and waits for two minutes. He then walks again. His journey takes 74 minutes.

Fig. 1.1

For the whole journey calculate:

the average speed.
average speed =

average speed = candidate's (i) / time
(average speed $=17 \times 60 / 74=$ ) $14 \mathrm{~km} / \mathrm{h}$

[Maximum number: 2]

A car accelerates uniformly in a straight line from rest at time t=0. At $t=3.2 \mathrm{~s}$ , the speed of the car is $13.0 \mathrm{~m} / \mathrm{s}$ .

The car travels at $13.0 \mathrm{~m} / \mathrm{s}$ from $t=3.2 \mathrm{~s}$ to $t=12.0 \mathrm{~s}$ .

Plot the speed-time graph for the car from t=0 to $t=12.0 \mathrm{~s}$ .

Figure for Question 6 — CAIE IGCSE Physics

[Maximum number: 1]

A car travels along a road at a constant speed.
What correctly describes the speed-time graph of the car?

A

a horizontal line

B

a vertical line

C

a diagonal line upwards

D

a curved line with increasing gradient

[Maximum number: 3]

A car of mass m is travelling along a straight, horizontal road at a constant speed v.
At time t=0, the driver of the car sees an obstruction in the road ahead of the car and applies the brakes.

The car does not begin to decelerate at t=0.

(a)

Fig. 1.1 is the distance-time graph for the car from t=0.

Fig. 1.1

[ 3 ]

(i)

State the property of a distance-time graph that corresponds to speed.

[ 1 ]

(ii)

Using Fig. 1.1, determine the initial speed v of the car.

v=

[ 2 ]

[Maximum number: 2]

A car accelerates uniformly in a straight line from rest at time t=0. At $t=3.2 \mathrm{~s}$ , the speed of the car is $13.0 \mathrm{~m} / \mathrm{s}$ .

The car travels at $13.0 \mathrm{~m} / \mathrm{s}$ from $t=3.2 \mathrm{~s}$ to $t=12.0 \mathrm{~s}$ .

Determine the distance travelled by the car between t=0 and $t=3.2 \mathrm{~s}$ .

[Maximum number: 3]

A student drops a heavy ball from a vertical height of 1.8 m above the ground. The ball then falls to the ground. It does not bounce after hitting the ground.

Calculate the maximum speed of the ball. Ignore air resistance.

Show your working.
maximum speed =

$5.9 \mathrm{~m} / \mathrm{s}$ $(\Delta) E_{\mathrm{p}}=(\Delta) E_{\mathrm{k}}$ OR $E_{\mathrm{p}}$ lost $=E_{\mathrm{k}}$ gained OR gravitational potential energy lost = kinetic energy gained $\mathbf{OR} m g(\Delta) h=1 / 2 m v^{2}$ $v^{2}=2 g(\Delta) h \mathbf{~ OR} v^{2}=2 \times 9.8 \times 1.8 \mathbf{OR} v^{2}=35(.28)$

[Maximum number: 1]

A ball of mass 130 g is launched from the ground at an initial velocity of $14 \mathrm{~m} / \mathrm{s}$ vertically upwards. It decelerates until it is at rest momentarily at a height h above the ground.

The acceleration of free fall is $9.8 \mathrm{~m} / \mathrm{s}^{2}$ .

Show that the time taken for the ball to reach height h is 1.4 s. Ignore the effect of air resistance.

$a=\Delta v /(\Delta) t \mathbf{A N D}(\Delta t=) 14 / 9.8 \mathbf{O R}(\Delta t=) \Delta v / a=14 / 9.8$

[Maximum number: 3]

Fig. 1.2 shows a vertical speed-time graph for a parachutist who jumps from a stationary hot-air balloon.

Fig. 1.2

The parachutist jumps from the balloon at time =0 and reaches the ground at B. The point A indicates when the parachute opens.

On Fig. 1.2, label a point on the graph where the acceleration is:
- zero with ' 1 '
- negative with ' 2 '
- decreasing with ' 3 '.

[Maximum number: 3]

A long tube contains oil. A small ball is held at rest at the surface of the oil. At time t=0, the ball is released and begins to fall vertically through the oil.

Fig. 1.1 shows the ball falling through the oil.

Fig. 1.1

As the ball begins to fall through the oil, it accelerates.

As the ball falls, its speed v is recorded. Fig. 1.2 is the speed-time graph for the falling ball.

Fig. 1.2

By drawing a tangent on Fig. 1.2, determine a value for the acceleration of the ball at $t=0.010 \mathrm{~s}$ .
acceleration =

tangent drawn at $t=0.010 \mathrm{~s}$ $1.2 \mathrm{~m} / \mathrm{s}^{2} \leqslant$ acceleration $\leqslant 1.8 \mathrm{~m} / \mathrm{s}^{2}$
( a= ) gradient of tangent $\mathbf{OR}(a=)\{\Delta y / \Delta x\}$

[Maximum number: 2]

A ball of mass 130 g is launched from the ground at an initial velocity of $14 \mathrm{~m} / \mathrm{s}$ vertically upwards. It decelerates until it is at rest momentarily at a height h above the ground.

Define deceleration.

(deceleration is) decrease in velocity per unit time OR rate of decrease in velocity OR negative rate of change of velocity
$\mathbf{O R}-\Delta v / \Delta t$
negative acceleration OR change in velocity per unit time OR rate of change of velocity

[Maximum number: 3]

A ball of mass 130 g is launched from the ground at an initial velocity of $14 \mathrm{~m} / \mathrm{s}$ vertically upwards. It decelerates until it is at rest momentarily at a height h above the ground.

The ball is dropped from the top of a tall building.

Describe and explain the motion of the ball as it falls. Consider the effect of air resistance in your answer.

any three from:

- (ball) accelerates

- (accelerates) at $9.8 \mathrm{~m} / \mathrm{s}^{2}$ initially OR (accelerates) due to force of gravity

- air resistance / resistive force increases (with speed / velocity)

- resultant force (downwards) decreases

- acceleration decreases

- terminal velocity is reached when acceleration is zero OR terminal velocity is reached when resultant force is zero

CAIE IGCSE Physics 1.2 Motion Topic Practice

Covered syllabus points

CAIE IGCSE Physics 1.2 Motion Topic Practice

Covered syllabus points

Question 1

Question 2

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Question (i)

Question (b)

Question (i)

Question 3

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Question (i)

Question (ii)

Question (b)

Question (c)

Question 4

Question 5

Question 6

Question 7

Question 8

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Question (i)

Question (ii)

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15