IB Physics SL A.1 Kinematics

[Maximum number: 4]

A ball of mass 0.250 kg is released from rest at time t=0, from a height H above a horizontal floor.

The graph shows the variation with time t of the velocity v of the ball. Air resistance is negligible. Take $g=-9.80 \mathrm{~ms}^{-2}$ . The ball reaches the floor after 1.0 s .

(a)

Determine H.

[ 1 ]

$H = \frac{1}{2} g t^{2} = 4.9\,\mathrm{m}$

Marking guidance:

Accept other methods as area from graph, alternative kinematics equations or conservation of mechanical energy.

Award [1] for a bald correct answer in the range 4.9-5.1

Award [0] if time used is different than 1.0 s

(b)

Label the time and velocity graph, using the letter M , the point where the ball reaches the maximum rebound height.

[ 1 ]

M at $1.6 \mathrm{~s}$

(c)

State the acceleration of the ball at the maximum rebound height.

[ 1 ]

$g = 9.80\,\mathrm{m}\,\mathrm{s}^{-2}$

Marking guidance:

Accept 9.81, 10 or a plain "g" Ignore sign if provided.

(d)

Draw, on the axes, a graph to show the variation with time of the height of the ball from the instant it rebounds from the floor until the instant it reaches the maximum rebound height. No numbers are required on the axes.

[ 1 ]

[Maximum number: 2]

A toy rocket is made from a plastic bottle that contains some water.

Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.

The air-water mixture is called the propellant.
The variation with time of the vertical velocity of the bottle is shown.

The bottle reaches its highest point at time $T_{1}$ on the graph and returns to the ground at time $T_{2}$ . The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.

Estimate the acceleration of the bottle when it is at its maximum height.

Attempt to calculate gradient of line at $t=1.2 \mathrm{~s}$

«-» 9.8 « $\mathrm{m} \mathrm{s}^{-2}$ » (accept 9.6-10.0)

[Maximum number: 8]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.
Part 1 Kinematics and gravitation
A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

(a)

State the significance of the negative values of s.

[ 1 ]

(b)

Use the graph to

[ 3 ]

(i)

estimate the velocity of the ball at $t=0.80 \mathrm{~s}$ .

[ 3 ]

tangent drawn to curve at 0.80 s ;
correct calculation of gradient of tangent drawn;
$-1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ or $1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ downwards;
or
correct coordinates used from the graph;
substitution into a correct equation;
$-1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ or $1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ downwards;

(c)

Calculate the speed of an identical ball when it falls 3.0 m from rest close to the surface of Earth. Ignore air resistance.

[ 1 ]

$\quad 7.7 \mathrm{~m} \mathrm{~s}^{-1}$ ;

(d)

Sketch, on the graph, the variation with time t of the displacement s from the point of release of the ball when the ball is dropped close to the surface of Earth. (For this sketch take the direction towards the Earth as being negative.)
Part 2 Radioactivity

Two isotopes of calcium are calcium-40 $\left({ }_{20}^{40} \mathrm{Ca}\right)$ and calcium-47 $\left({ }_{20}^{47} \mathrm{Ca}\right)$ . Calcium-40 is stable and calcium-47 is radioactive with a half-life of 4.5 days.

[ 3 ]

Curve permanently below Moon curve.
Smooth parabola (judge by eye).
Line passing through $s=-3.00\ \text{m}, t=0.78\ \text{s}$ or $s=-3.50\ \text{m}, t=0.84\ \text{s}$ ( $\pm 1\ \text{mm}$ ).

Figure for Answer to Question (d) — IB Physics SL

[Maximum number: 1]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.
Part 1 Kinematics and gravitation
A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

State the significance of the negative values of s.

[Maximum number: 1]

The road from city X to city Y is 1000 km long. The displacement is 800 km from X to Y.

What is the distance travelled from Y to X and the displacement from Y to X ?

Distance travelled

from Y to X / km

Displacement from

Y to X / km

800

1000

800

-800

1000

-800

[Maximum number: 3]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.
Part 1 Kinematics and gravitation
A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

Use the graph to

estimate the velocity of the ball at $t=0.80 \mathrm{~s}$ .

tangent drawn to curve at 0.80 s ;
correct calculation of gradient of tangent drawn;
$-1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ or $1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ downwards;
or
correct coordinates used from the graph;
substitution into a correct equation;
$-1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ or $1.3 \pm 0.1 \mathrm{~m} \mathrm{~s}^{-1}$ downwards;

[Maximum number: 2]

This question is in two parts. Part 1 is about the motion of a ship. Part 2 is about melting ice.

The ship's engines are switched off and the ship comes to rest from a speed of $7 \mathrm{~m} \mathrm{~s}^{-1}$ in a time of 650 s .

Estimate the distance that the ship takes to stop. Assume that the acceleration is uniform.

[Maximum number: 2]

This question is in two parts. Part 1 is about the motion of a ship. Part 2 is about melting ice.

The ship's engines are switched off and the ship comes to rest from a speed of $7 \mathrm{~m} \mathrm{~s}^{-1}$ in a time of 650 s .

It is unlikely that the acceleration of the ship will be uniform given that the resistive force acting on the ship depends on the speed of the ship. Using the axes, sketch a graph to show how the speed v varies with time t after the ship's engines are switched off.

starts at $7.0 / 7.2 \mathrm{~m} \mathrm{~s}^{-1}$ ;
(allow ECF from (d)(i)) correct shape;

Figure for Answer to Question 8 — IB Physics SL

[Maximum number: 5]

A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of $64 \mathrm{~m} \mathrm{~s}^{-1}$ . The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms .

(a)

The student strikes the tennis ball at point P . The tennis ball is initially directed at an angle of $7.00^{\circ}$ to the horizontal.

The following data are available.

[ 5 ]

(i)

Calculate the time it takes the tennis ball to reach the net.

[ 2 ]

horizontal component of velocity is $64.0 \times \cos 7^{\circ}=63.52<\mathrm{ms}^{-1} \gg$ $t=<\frac{11.9}{63.52}=>0.187 / 0.19<\mathrm{s} \gg$

Marking guidance:

Do not award BCA. Check working. Do not award ECF from using $64 \mathrm{~m} \mathrm{~s}^{-1}$ .

(ii)

Show that the tennis ball passes over the net.

[ 3 ]

ALTERNATIVE 1
$u_{y}=64 \sin 7 / 7.80<\mathrm{ms}^{-1} \gg$

decrease in height $=7.80 \times 0.187+\frac{1}{2} \times 9.81 \times 0.187^{2} / 1.63$ «m»

final height = « 2.80-1.63 » =1.1 / 1.2 «m»

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «=2.80-0.91 »=1.89 «m»

time to fall this distance found using $<1.89=7.8 t+\frac{1}{2} \times 9.81 \times t^{2}$ »

t=0.21 «s»

0.21 «s» > 0.187 «s»

«reaches the net before it has fallen far enough so goes over»

Other alternatives are possible

[Maximum number: 3]

This question is about kinematics.

After the stone in (a) hits the water surface it rapidly reaches a terminal speed as it falls through the water. The stone leaves Fiona's hand at time t=0. It hits the water surface at $t_{1}$ and it comes to rest at the bottom of the water at $t_{2}$ . Using the axes below, sketch a graph to show how the speed v of the stone varies from time t=0 to just before $t=t_{2}$ . (There is no need to add any values to the axes.)

Figure for Answer to Question 10 — IB Physics SL

straight line to water surface;
clear decrease after hitting surface;
constant non-zero speed reached smaller than maximum; $; \begin{aligned} & \text { (speed must be less than } \\ & \text { maximum velocity) }\end{aligned}$

IB Physics SL A 1 Kinematics Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question 3

Question (a)

Question (b)

Question (i)

Question (c)

Question (d)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question (a)

Question (i)

Question (ii)

Question 10