# IBDP Physics Unit 1. Measurements and uncertainties: IB Style Questions- SL Paper 2

## Question

A small ball of mass is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl. The normal reaction force makes an angle θ to the horizontal.

State the direction of the resultant force on the ball.


a.i.

On the diagram, construct an arrow of the correct length to represent the weight of the ball. a.ii.

Show that the magnitude of the net force on the ball is given by the following equation.

F=mg/tanθ


a.ii   i    .

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.


b.

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.


c.

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m. The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.


d.

## Markscheme

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

a.i.

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required [2 marks]

a.ii.

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with Fand W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship a.iii. Award  for a bald correct answer

Award  for an answer of 13.9/14 «ms 1». MP2 omitted

[4 marks]

b.

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

c. Allow 12.5 from incorrect use of kinematics equations

Award  for a bald correct answer

Award  for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

d.

## Question

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. The glider reaches its launch speed of 27.0 m 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.


a.

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.


b.

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


c.

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 m s–1. Include an appropriate unit for your answer.


d.

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.


e.

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


f.

At a particular instant in the flight the glider is losing 1.00 m of vertical height for every 6.00 m that it goes forward horizontally. At this instant, the horizontal speed of the glider is 12.5 m s–1. Calculate the velocity of the glider. Give your answer to an appropriate number of significant figures.


g.

## Markscheme

correct use of kinematic equation/equations

148.5 or 149 or 150 «m»

Substitution(s) must be correct.

a.

a =

$\frac{27/}{11}$

or 2.45 «m s–2»

F – 160 = 492 × 2.45

1370 «N»

Could be seen in part (a).
Award  for solution that uses a = 9.81 m s–2

b.

ALTERNATIVE 1

«work done to launch glider» = 1370 x 149 «= 204 kJ»

«work done by motor»

$=\left(\frac{204×100\right)/}{23}$

«power input to motor»

 $=\left(\frac{204×100\right)/}{23}×\frac{1/}{11}=80$

or 80.4 or 81 k«W»

ALTERNATIVE 2

use of average speed 13.5 m s–1

«useful power output» = force x average speed «= 1370 x 13.5»

power input = «

“>

» 80 or 80.4 or 81 k«W»

ALTERNATIVE 3

work required from motor = KE + work done against friction «

“>» = 204 «kJ»

«energy input

= work required from motor *100/23

power input

“>

k«W»

Award [2 max] for an answer of 160 k«W».

c.

«

“>

Do not accept Hz.
Award [1 max] if unit is missing.

d. drag correctly labelled and in correct direction

weight correctly labelled and in correct direction AND no other incorrect force shown

Award [1 max] if forces do not touch the dot, but are otherwise OK.

e.

name Newton’s first law

vertical/all forces are in equilibrium/balanced/add to zero
OR
vertical component of lift mentioned

as equal to weight

f.

any speed and any direction quoted together as the answer

quotes their answer(s) to 3 significant figures

speed = 12.7 m s–1 or direction = 9.46º or 0.165 rad «below the horizontal» or gradient of

g