IB Physics HL A.2 Forces and Momentum

[Maximum number: 7]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

(a)

On the diagram, draw and label one other force acting on the skier.

[ 1 ]

arrow vertically downwards labelled weight/ W / mg / gravitational
force $/ \mathrm{F}_{\mathrm{g}} / \mathrm{F}_{\text {gravitational }} /$ force of gravity;
(judge by eye)

Do not allow "gravity".

(b)

Calculate the magnitude of the normal reaction acting on the skier.

[ 2 ]

$\quad(N=) m g \cos \theta /$ correct substitution;
$\left(=73 \times 9.81 \times \cos 12^{\circ}=\right) 700 \mathrm{~N}$ ;

(c)

The total frictional force acting is 65 N . Determine the tension in the rope.

[ 2 ]

tension = frictional force + component of weight parallel to slope /
tension $=65+m g \sin \theta$ ;
$214 / 210 \mathrm{~N}$ ;

(d)

Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.

[ 2 ]

[Maximum number: 15]

A girl on a sledge is moving down a snow slope at a uniform speed.

(a)

Draw the free-body diagram for the sledge at the position shown on the snow slope.

[ 2 ]

arrow vertically downwards labelled weight/gravitational force $F_{\mathrm g}$ ;
arrow perpendicular to the snow slope labelled reaction/normal contact force $F_{\mathrm N}$ ;
friction force acting up the slope

(b)

After leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.

[ 3 ]

(c)

When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg . Calculate the speed of the sledge immediately before the girl jumps from it.

[ 2 ]

mention of conservation of momentum

OR

$5.5\times4.2=(55+5.5)v$ $v=0.38\ \mathrm{m\,s^{-1}}$

(d)

The girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.

[ 3 ]

same change in momentum/impulse;
the time taken to stop would be greater with the snow

$F=\frac{\Delta p}{\Delta t}$ , therefore F is smaller with the snow

OR

force is proportional to rate of change of momentum, therefore F is smaller with the snow

(e)

The sledge, without the girl on it, now travels up a snow slope that makes an angle of $6.5^{\circ}$ to the horizontal. At the start of the slope, the speed of the sledge is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The coefficient of dynamic friction of the sledge on the snow is 0.11 .

[ 3 ]

(i)

Show that the acceleration of the sledge is about $-2 \mathrm{~m} \mathrm{~s}^{-2}$ .

[ 3 ]

friction force down slope $=\mu mg\cos(6.5^\circ)=5.9\ \mathrm N$

component of weight down slope $=mg\sin(6.5^\circ)=6.1\ \mathrm N$

acceleration $=\frac{12}{5.5}=2.2\ \mathrm{m\,s^{-2}}$

(f)

The coefficient of static friction between the sledge and the snow is 0.14 . Outline, with a calculation, the subsequent motion of the sledge.

[ 2 ]

maximum frictional force $=\mu R=7.5\ \mathrm N$ ;
the sledge will not move because the maximum static friction force is greater than the component of weight down the slope

[Maximum number: 6]

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

(a)

The thread makes an angle of $30^{\circ}$ with the vertical wall. The ball has a mass of 0.025 kg .

Determine the horizontal force that acts on the ball.

[ 3 ]

Horizontal force on ball: $F = T\sin 30$ .
$T = \frac{mg}{\cos 30}$ .
$F = mg\tan 30 = 0.025 \times 9.8 \times \tan 30 = 0.14\ \text{N}$ .

Allow $g=10\ \text{N kg}^{-1}$ .
Award [3] marks for a bald correct answer.
Award [1 max] for an answer of zero if the horizontal force is interpreted as the horizontal component of the net force.

(b)

The thread breaks. Explain the initial subsequent motion of the ball.

[ 3 ]

horizontal/repulsive force and vertical force/pull of gravity act on the ball so ball has constant acceleration/constant net force motion is in a straight line at $30^{\circ}$ to vertical away from wall/along original line of thread

3 max

[Maximum number: 2]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.

[Maximum number: 3]

A girl on a sledge is moving down a snow slope at a uniform speed.

After leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.

[Maximum number: 2]

A girl on a sledge is moving down a snow slope at a uniform speed.

Draw the free-body diagram for the sledge at the position shown on the snow slope.

arrow vertically downwards labelled weight/gravitational force $F_{\mathrm g}$ ;
arrow perpendicular to the snow slope labelled reaction/normal contact force $F_{\mathrm N}$ ;
friction force acting up the slope

[Maximum number: 3]

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

The thread makes an angle of $30^{\circ}$ with the vertical wall. The ball has a mass of 0.025 kg .

Determine the horizontal force that acts on the ball.

Horizontal force on ball: $F = T\sin 30$ .
$T = \frac{mg}{\cos 30}$ .
$F = mg\tan 30 = 0.025 \times 9.8 \times \tan 30 = 0.14\ \text{N}$ .

Allow $g=10\ \text{N kg}^{-1}$ .
Award [3] marks for a bald correct answer.
Award [1 max] for an answer of zero if the horizontal force is interpreted as the horizontal component of the net force.

[Maximum number: 1]

The diagram shows two blocks of mass m and M in contact on a frictionless surface. A force F is applied on the block of mass m and causes the blocks to move with an acceleration a.

What is the force that the block of mass M is exerting on the block of mass m ?

A

m a

B

Ma

C

(M+m) a

D

$\frac{m}{M+m} a$

[Maximum number: 2]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

Calculate the magnitude of the normal reaction acting on the skier.

$\quad(N=) m g \cos \theta /$ correct substitution;
$\left(=73 \times 9.81 \times \cos 12^{\circ}=\right) 700 \mathrm{~N}$ ;

[Maximum number: 2]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

The total frictional force acting is 65 N . Determine the tension in the rope.

tension = frictional force + component of weight parallel to slope /
tension $=65+m g \sin \theta$ ;
$214 / 210 \mathrm{~N}$ ;

[Maximum number: 2]

An object attached to a spring oscillates with simple harmonic motion in a horizontal plane. The graph shows how the kinetic energy of the object varies with the displacement x.

The mass of the object is 0.15 kg .

Calculate the maximum force exerted on the object by the spring.

ALTERNATIVE 1
$\begin{aligned} & a_{\max }=8.6^{2} \times 0.060 \text { OR }\left(\frac{2 \pi}{0.73}\right)^{2} \times 0.060<=4.44 \mathrm{~m} \mathrm{~s}^{-2}> & F_{\max }=0.15 \times 4.44=0.67<\mathrm{N} \gg \end{aligned}$

ALTERNATIVE 2
$\begin{aligned} & k=8.6^{2} \times 0.15 \text { OR }\left(\frac{2 \pi}{0.73}\right)^{2} \times 0.15<=11.1 \mathrm{~N} \mathrm{~m}^{-1}> & F_{\max }=11.1 \times 0.060=0.67<\mathrm{N}> & \end{aligned}$

MP1 find a or k MP2 F=m a / F=k x

Award [2] for BCA

[Maximum number: 4]

A solid sphere is released from rest below the surface of a fluid and begins to fall.

(a)

Explain why the sphere will reach a terminal speed.

[ 2 ]

(b)

The weight of the sphere is 6.16 mN and the radius is $5.00 \times 10^{-3} \mathrm{~m}$ . For a fluid of density $8.50 \times 10^{2} \mathrm{~kg} \mathrm{~m}^{-3}$ , the terminal speed is found to be $0.280 \mathrm{~ms}^{-1}$ . Calculate the viscosity of the fluid.

[ 2 ]

any evidence (numerical or algebraic) of a realisation that
$\begin{aligned}6\pi\eta rv+\rho gV&=W \\ \eta&=\frac{6.16\times10^{-3}-4.366\times10^{-3}}{6\pi\times0.005\times0.280} \\ &=0.0680\ \mathrm{Pa\,s}\end{aligned}$

[Maximum number: 1]

A group of students is trying to determine the density and the viscosity of a liquid.

To determine the density, they use a balance to read the mass m of a sphere in air and immersed in the liquid.

They use a sphere of volume $V=1.827 \times 10^{-7} \mathrm{~m}^{3}$ .
The readings are $m_{\text {air }}=1.427 \mathrm{~g}$ in air and $m_{\text {immersed }}=1.208 \mathrm{~g}$ in the liquid.
The readings are different due to buoyancy. The buoyancy force $F_{\mathrm{b}}$ is given by

F_{\mathrm{b}}=\rho V g

where V is the volume of the sphere and $\rho$ is the density of the liquid.

Show that $F_{\mathrm{b}}$ is about 2 mN .

$(0.001427-0.001208) \cdot 9.8$ OR 2.1 mN

Allow the calculation to give a result
in mN or N . Look for full substitution
or answer with at least two SF.

[Maximum number: 1]

Which new concept was required to understand action at a distance?

A

Wave motion

B

Atomic theory

C

Tunnelling

D

Fields

[Maximum number: 1]

A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is 16 V and it can deliver a charge of 43 kC when discharging completely from a full charge.

The bicycle and the girl have a total mass of 66 kg . The girl rides up a slope that is at an angle of $3.0^{\circ}$ to the horizontal.

Calculate the component of weight for the bicycle and girl acting down the slope.

$66 \mathrm{~g} \sin \left(3^{\circ}\right)=34$ « N »

[Maximum number: 1]

This question is in two parts

The magnitude of gravitational field strength g is defined from the equation shown below.

g=\frac{F_{\mathrm{g}}}{m}

The magnitude of electric field strength E is defined from the equation shown below.

E=\frac{F_{\mathrm{E}}}{q}

For each of these defining equations, state the meaning of the symbols

$F_{\mathrm{E}}$ .

[Maximum number: 2]

A girl on a sledge is moving down a snow slope at a uniform speed.

When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg . Calculate the speed of the sledge immediately before the girl jumps from it.

mention of conservation of momentum

OR

$5.5\times4.2=(55+5.5)v$ $v=0.38\ \mathrm{m\,s^{-1}}$

[Maximum number: 3]

A girl on a sledge is moving down a snow slope at a uniform speed.

The girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.

same change in momentum/impulse;
the time taken to stop would be greater with the snow

$F=\frac{\Delta p}{\Delta t}$ , therefore F is smaller with the snow

OR

force is proportional to rate of change of momentum, therefore F is smaller with the snow

[Maximum number: 3]

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.

The bottle bounces when it returns to the ground.

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.

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

Force $=4.6+0.3=4.9\ \mathrm{N}$

Any answer to 2 sf

Marking guidance:

Accept ± 0.5 velocity values from graph

[Maximum number: 2]

One end of a light spring is attached to a rigid horizontal support.

An object W of mass 0.15 kg is suspended from the other end of the spring. The extension x of the spring is proportional to the force F causing the extension. The force per unit extension of the spring k is $18 \mathrm{Nm}^{-1}$ .

A student pulls W down such that the extension of the spring increases by 0.040 m . The student releases W and as a result W performs simple harmonic motion (SHM).

Determine the maximum acceleration of W .

$F=(18 \times 0.04=) 0.72 \mathrm{~N}$ ;
acceleration $=\frac{0.72}{0.15}=4.8 \mathrm{~m} \mathrm{~s}^{-2}$ ;

[Maximum number: 2]

This question is in two parts. Part 1 is about momentum and energy. Part 2 is about the de Broglie hypothesis and radioactive decay.

Jane and Joe are two ice skaters initially at rest on a horizontal skating rink. They are facing each other and Jane is holding a ball. Jane throws the ball to Joe who catches it. The speed at which the ball leaves Jane, measured relative to the ground, is $8.0 \mathrm{~ms}^{-1}$ . The following data are available.

\begin{aligned} & \text { Mass of Jane }=52 \mathrm{~kg} \\ & \text { Mass of Joe }=74 \mathrm{~kg} \\ & \text { Mass of ball }=1.3 \mathrm{~kg} \end{aligned}

Use the data to calculate the

speed V of Joe relative to the ground immediately after he catches the ball.

[Maximum number: 3]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

Determine in terms of M and v the energy lost during the collision.

$\text { initial } K E=\frac{M v^{2}}{6}$

final $K E=\frac{M v^{2}}{24}$

energy loss $=\frac{M v^{2}}{8}$

[Maximum number: 7]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

(a)

Determine the magnitude of the velocity of Aibhe relative to

[ 2 ]

(i)

Euan.

[ 1 ]

2.0 or $0\left(\mathrm{~ms}^{-1}\right)$ ;

(ii)

the centre of the merry-go-round.

[ 1 ]

1.0 or $0\left(\mathrm{~ms}^{-1}\right)$ ;

(b)

Outline why Aibhe is accelerating even though she is moving at constant speed.

[ 2 ]

(c)

Draw an arrow on the diagram on page 29 to show the direction in which Aibhe is accelerating.

[ 1 ]

(d)

Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below.

Euan pushes the merry-go-round so that he is again moving at $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

[ 2 ]

(i)

Calculate the magnitude of Aibhe's acceleration.
Part 2 Charge-coupled devices (CCDs)

[ 2 ]

$\quad a\left(=\frac{v^{2}}{r}\right)=\frac{0.5^{2}}{0.75}$ ;
$a=0.33\left(\mathrm{~ms}^{-2}\right)$ ;

Marking guidance:

Allow ECF from (d)(i).
Award [2] for a bald correct answer.

Part 2 Charge-coupled devices (CCDs)

[Maximum number: 1]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

Identify the force that is causing Aibhe to move in a circle.

[Maximum number: 2]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

The diagram below shows a side view of Aibhe and Euan on the merry-go-round.

Explain why Aibhe feels as if her upper body is being "thrown outwards", away from the centre of the merry-go-round.

[Maximum number: 3]

Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground.

Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below.

Euan pushes the merry-go-round so that he is again moving at $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

Determine Aibhe's speed relative to the ground.

Aibhe's period of revolution is the same as before;
from $v=\frac{2 \pi r}{T}$ , since r is halved, v is halved;
$v=0.5\left(\mathrm{~ms}^{-1}\right) ;$

Marking guidance:

Award [3] for a bald correct answer.

IB Physics HL A 2 Forces and Momentum Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (i)

Question (f)

Question 3

Question (a)

Question (b)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question (a)

Question (b)

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

Question (i)

Question 24

Question 25

Question 26