IB Physics SL A Space, Time & Motion

[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.

Figure for Question A. Space, time and motion question 1 — IB Physics SL

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: 12]

(a)

B3. This question is in two parts. Part 1 is about a collision. Part 2 is about electric current and resistance.
Part 1 A collision
Two identical blocks of mass 0.17 kg and length 0.050 m are travelling towards each other along a straight line through their centres as shown below. Assume that the surface is frictionless.

The initial distance between the centres of the blocks is 0.900 m and both blocks are moving at a speed of $0.18 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the surface.

[ 12 ]

(i)

Determine the time taken for the blocks to come into contact with each other.

[ 3 ]

(ii)

As a result of the collision, the blocks reverse their direction of motion and travel at the same speed as each other. During the collision, 20 % of the kinetic energy of the blocks is given off as thermal energy to the surroundings.

[ 5 ]

(i)

State and explain whether the collision is elastic or inelastic.

[ 2 ]

(ii)

Show that the final speed of the blocks relative to the surface is $0.16 \mathrm{~m} \mathrm{~s}^{-1}$ .

[ 3 ]

(iii)

State Newton's third law of motion.

[ 1 ]

(iv)

During the collision of the blocks, the magnitude of the force that block A exerts on block B is $F_{\mathrm{AB}}$ and the magnitude of the force that block B exerts on block A is $F_{\mathrm{BA}}$ . On the diagram below, draw labelled arrows to represent the magnitude and direction of the forces $F_{\mathrm{AB}}$ and $F_{\mathrm{BA}}$ .

[ 3 ]

[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 .
Part 2 Thermal physics

[ 3 ]

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

Part 2 Thermal physics

(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.

Figure for Question A. Space, time and motion question 4 — IB Physics SL

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 ]

IB Physics SL A: Space, Time and Motion

Covered syllabus points

Covered syllabus points

A. Space, time and motion question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

A. Space, time and motion question 2

Question (a)

Question (i)

Question (ii)

Question (i)

Question (ii)

Question (iii)

Question (iv)

A. Space, time and motion question 3

Question (a)

Question (b)

Question (i)

A. Space, time and motion question 4

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)