Exam Style Answers 16 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 16 c

Exam-style questions 1

C 

[1]

2

B 

[1]

3

a

b

  4

a

At the position shown, the weight of the truck  v2 speed spe ed of thetruck: thetruck: g  =   r v = gr = 9.8 × 4   v = 6.3 m s 1 

 

   

b  

=

  20   60 

 0.039 N 

   

The toy falls off because the frictional

Weight acting vertically downwards, lift force perpendicular to the aeroplane wings  [2]

b

(lift = L, angle with horizontal = θ)

 

Therefore the vertical force = L cos θ = mg   [1]

 

horizontal force = L sin θ = mv2/r 

 

Now, sinθ

change in potential energy = kinetic energy  1 mg ∆h = mv 2 ,  therefore, 2 2 m × 9.8× 0.70 m

centripetalforce F c

=

=

 

 0.46 N 

=

3.7 m s 1    

 

mv2 r

=

[1]

0.05 050 × 3.7 2 1.5

  [1] [1]

tension in the string T = F c+ mg = 0.46 +  ×

a

m × 752

mv2 =

tanθ =

r mg 

800 m × g 

=

b

=

  0.717 [1]

θ = 36° 

[1]

Angle covered per unit time 

[1]

Angle in radians 

 

[1]

i

T  cos  cos θ = mg  leading  leading to  T  =

[1] 0. 20 20 × 9. 8 cos 56°

 

[1]

[1]

−

cosθ

9

[1]

a

2

[1]

 

 

T = 3.5 

ii 

entripetalforce T sinθ   =

=

 

   iii iii

[1]

T sin θ

mr 6.0 rad s 1 

=

=

3.5 × sin θ 

mrω 2  

[1]

 

[1]

0.20 20 × 0.40 40 [1]

−

time period T   =

2π ω 

=

2π 6.0

=

1.0 s  

[1]

=

(0.05  9.8)  0.95 N 

1

[1]

The centripetal force on the toy t oy increases with its speed.  [1]

v= b

Friction between the tyres and the road provides provi des the centripetal force force.. 

[1]

force and the toy is notbetween sufficientthe to turntable provide the centripetal force.   [1] a

8

force = mrω 2 = 60 × 10−3 × 0.15 ×  2 × π

 

[1]

is not enough to push the car around the desired curve.) 

[1]

−

a

[1]

The ball is not in equilibrium in the vertical position because it has an upward (centripetal) acceleration. 

Oil  O il reduces the frictional force, so the car carries on in a wider arc. (The frictional force

[1]

[2]

6

7

−

b

5

The angle subtended at the centre of a circle by an arc equal in length to the radius.  [1] 2π angular angul ar speed ω   = [1]   15 1 = 0.42 rad s   [1]

 

The weight is only equal to the tension when the ball is at rest in the vertical position.  [1]

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

10 a

  b

 

The force acting towards the centre of a circle  [1] on a body moving round that circle   mv2 i   entripetalforce = mg  = r  

therefore, v = rg  =

 

v = 3.0 m s−1 

ii

F = 2mg  = 2 × 5.4 × 9.8 

 

=

×

9.8  

2

[1]

Inertia means that with no force pilot would travel in a straight line.   Centripetal force pulls the pilot in towards the centre of the circle.  mv 2 F  = i , therefore, r Fr

6 m × 9.8 × 5  

=

m m s−1  v = 17.1    

   ii ii

   

86 rad s  

[1]

angle in radians = length of arc radius

 

For a complet completee circle, arc length =  circumference = 2pr 

[1]

angle in radians = 2pr/r = 2π 

[1]

  b

[1] [1]

c

 

i

540 rpm = 540/60 rev s−1, therefore, ω  = 9 × 2π rad s−1  [1]

 

v = rω  = 0.10 × 9 × 2π = 5.7 rad s−1  [1]

 ii ii

F

 

=

=

mrω 2

=

20 ×10

6

−

×

0.10 × 5.72  

 6.4 × 10 3 N 

[1] [1]

−

gravitational force ≈ 20 × 10  

3

−

×

 10 = 0.2 N [1]

Very much less than the centripetal force 

[1]

[2]

m

[1]

ω  = v/r = 17.1 × 5  =

mean the pilot will experience high g   forces. 

[1]

 

v=

  12 a

[1]

Pilot feels thrown out from the circle or pilot feels pressed outwards against seat. [1]

b

[1]

[1] [1]

 106 N 

Sharp turns in combat at high speed  

[1]

11 a

 

2

1.8

c

[1]

1

−

revs per minute = 86 × 60/2 p = 820 revs per minute 

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020