IB Physics SL A Space Time and Motion Questions

Practise IB Physics SL mechanics through kinematics, forces, energy, momentum and rotation, using equations, diagrams and data to justify each step.

Exam points

Apply kinematics, force, momentum, work, energy and power relationships to motion problems and data.
Analyse rigid-body rotation, torque, angular momentum and energy transfer using SL models.
Use Galilean and special-relativity concepts to interpret frames, time, length and spacetime evidence.

Covered syllabus points

A.1 Kinematics A.2 Forces and momentum A.3 Work, energy and power

[Maximum number: 9]

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 ]

$H=\ll \frac{1}{2} g t^{2} \Rightarrow>4.9 \ll \mathrm{~m} \gg$

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 ]

« $g=>9.80<\mathrm{ms}^{-2} \gg$

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 ]

(e)

Estimate the loss in the mechanical energy of the ball as a result of the collision with the floor.

[ 1 ]

«loss of KE is $\frac{1}{2} \times 0.25 \times\left(9.8^{2}-5^{2}\right)=>8.9$ «J»

Marking guidance:

Award [1] mark for an answer in the range 8.7-9.5

(f)

Determine the average force exerted on the floor by the ball.

[ 3 ]

$\begin{aligned} \Delta p=0.250 \times(9.8+5.0) F_{\text {net }}=<\frac{\Delta p}{\Delta t}=\frac{3.7}{0.1}=>37 \ll \mathrm{~N} \gg N=37+0.250 \times 9.8=39.5 \ll \mathrm{~N} \gg \end{aligned}$

Allow ECF for MP2 and MP3

(g)

Suggest why the momentum of the ball was not conserved during the collision with the floor.

[ 1 ]

[Maximum number: 14]

A glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from rest by a cable pulled by a motor that exerts a horizontal force on the glider throughout the launch.

(a)

The glider reaches its launch speed of $27.0 \mathrm{~m} \mathrm{~s}^{-1}$ after accelerating for 11.0 s . Assume that the glider moves horizontally until it leaves the ground. Calculate the total distance travelled by the glider before it leaves the ground.

[ 2 ]

(b)

The glider and pilot have a total mass of 492 kg . During the acceleration the glider is subject to an average resistive force of 160 N . Determine the average tension in the cable as the glider accelerates.

[ 3 ]

$a=\frac{27}{11}=2.45\,\mathrm{m\,s^{-2}}$ $F-160\,\mathrm{N}=492\,\mathrm{kg}\times2.45\,\mathrm{m\,s^{-2}}$ $F=1370\,\mathrm{N}$

This calculation could also be shown in part (a).

Award [0] for a solution that uses $a=9.81\,\mathrm{m\,s^{-2}}$ .

(c)

The cable is pulled by an electric motor. The motor has an overall efficiency of 23 %. Determine the average power input to the motor.

[ 3 ]

Alternative 1:
$W_{\text{launch}}=1370\,\mathrm{N}\times149\,\mathrm{m}=204\,\mathrm{kJ}$ $W_{\text{motor}}=W_{\text{launch}}\times\frac{100}{23}$ $P_{\text{input}}=\frac{204\,\mathrm{kJ}\times100/23}{11\,\mathrm{s}}=80.4\,\mathrm{kW}$

Alternative 2:
$v_{\text{average}}=13.5\,\mathrm{m\,s^{-1}}$ $P_{\text{useful}}=1370\,\mathrm{N}\times13.5\,\mathrm{m\,s^{-1}}$ $P_{\text{input}}=1370\times13.5\times\frac{100}{23}=80.4\,\mathrm{kW}$

Alternative 3:
$W_{\text{motor}}=\frac12(492\,\mathrm{kg})(27\,\mathrm{m\,s^{-1}})^2+(160\,\mathrm{N})(148.5\,\mathrm{m})=204\,\mathrm{kJ}$ $E_{\text{input}}=W_{\text{motor}}\times\frac{100}{23}$ $P_{\text{input}}=\frac{883000\,\mathrm{J}}{11\,\mathrm{s}}=80.3\,\mathrm{kW}$

Award [2 max] for an answer of 160 kW.

(d)

The cable is wound onto a cylinder of diameter 1.2 m . Calculate the angular velocity of the cylinder at the instant when the glider has a speed of $27 \mathrm{~m} \mathrm{~s}^{-1}$ . Include an appropriate unit for your answer.

[ 2 ]

\omega=\frac{v}{r}=\frac{27\,\mathrm{m\,s^{-1}}}{0.6\,\mathrm{m}}=45\,\mathrm{rad\,s^{-1}}

Do not accept Hz.

Award [1 max] if the unit is missing.

(e)

After takeoff the cable is released and the unpowered glider moves horizontally at constant speed. The wings of the glider provide a lift force. The diagram shows the lift force acting on the glider and the direction of motion of the glider.

Draw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and vertical directions.

[ 2 ]

(f)

Explain, using appropriate laws of motion, how the forces acting on the glider maintain it in level flight.

[ 2 ]

[Maximum number: 5]

This question is in two parts. Part 1 is about energy resources. Part 2 is about thermal physics.
Part 1 Energy resources
Electricity can be generated using nuclear fission, by burning fossil fuels or using pump storage hydroelectric schemes.

(a)

A hydroelectric scheme has an efficiency of 92 %. Water stored in the dam falls through an average height of 57 m . Determine the rate of flow of water, in $\mathrm{kg} \mathrm{s}^{-1}$ , required to generate an electrical output power of 4.5 MW .

[ 3 ]

use of $\frac{m g h}{t}$ ;
$\frac{m}{t}=\frac{4.5 \times 10^{6}}{0.92 \times 9.81 \times 57}$ ;
$8.7 \times 10^{3} \mathrm{~kg} \mathrm{~s}^{-1}$ ;

(b)

A mass of 0.22 kg of lead spheres is placed in a well-insulated tube. The tube is turned upside down several times so that the spheres fall through an average height of 0.45 m each time the tube is turned. The temperature of the spheres is found to increase by $8^{\circ} \mathrm{C}$ .

[ 2 ]

(i)

Discuss the changes to the energy of the lead spheres.

[ 2 ]

[Maximum number: 12]

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.

(a)

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

[ 2 ]

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

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

(b)

The bottle bounces when it returns to the ground.

[ 5 ]

(i)

Calculate the fraction of the kinetic energy of the bottle that remains after the bounce.

[ 2 ]

Attempt to evaluate KE ratio as $\left(\frac{v_{\text {final }}}{v_{\text {initial }}}\right)^{2}$

« $\left(\frac{4.5}{10}\right)^{2}=$ 》 0.20 OR 20 % OR $\frac{1}{5}$

Marking guidance:

Accept ± 0.5 velocity values from graph

(ii)

The mass of the bottle is 27 g and it is in contact with the ground for 85 ms .

Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.

[ 3 ]

Attempt to use force = momentum change ÷ time $\ll=\frac{(4.5+10) \times 0.027}{85 \times 10^{-3}}=4.6 \gg$

Force = «4.6 + 0.3» 4.9 «N»

Any answer to 2 sf

Marking guidance:

Accept ± 0.5 velocity values from graph

(c)

After a second bounce, the bottle rotates about its centre of mass. The bottle rotates at 0.35 revolutions per second.

The centre of mass of the bottle is halfway between the base and the top of the bottle. Assume that the velocity of the centre of mass is zero.

Calculate the linear speed of the top of the bottle.

[ 3 ]

Alternative 1:
$\omega=2\pi(0.35)=2.20\ \mathrm{rad\ s^{-1}}$ .
Use $v=0.14\omega$ .
$v=0.31\ \mathrm{m\ s^{-1}}$ .

Alternative 2:
$T=1/0.35=2.9\ \mathrm{s}$ .
$v=2\pi(0.14)/T=0.31\ \mathrm{m\ s^{-1}}$ .

Award [3] for BCA.

\omega=2\pi(0.35)=2.20\ \mathrm{rad\ s^{-1}},\qquad v=0.14\omega=0.31\ \mathrm{m\ s^{-1}}

(d)

The maximum height reached by the bottle is greater with an air-water mixture than with only high-pressure air in the bottle.

Assume that the speed at which the propellant leaves the bottle is the same in both cases.

Explain why the bottle reaches a greater maximum height with an air-water mixture.

[ 2 ]

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question 3

Question (a)

Question (b)

Question (i)

Question 4

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)