• 9702 November 2005 Paper 2 Worked Solutions | A-Level Physics

Paper 2

Question 1

(a)

(i)

Pressure is defined as the force per unit area

(ii)

Base units of pressure:

kgm^-1s^-2

(b)

Pressure p at depth h in an incompressible fluid of density ρ is given by

p = ρgh            where g is acceleration of free fall

Use base units to check homogeneity of equation:

Base unit of ρ: kgm^-3

Base unit of g: ms^-2

So, base unit of hρg: (m)(kgm^-3)(ms^-2) = (kgm^-1s^-2)

This is the same unit as pressure

Question 2

{Detailed explanations for this question is available as Solution 669 at Physics 9702 Doubts | Help Page 134 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-134.html}

  

Question 3

Stone on a string is made to travel along horizontal circular path, as shown. Stone has constant speed.

(a)

Acceleration is defined as the change in velocity per time (taken)

(b)

Explain whether stone is accelerating:

Velocity is a vector / Velocity has both a magnitude and a direction. Since the direction is changing, the stone must be acceleration.

(c)

Stone has weight of 5.0N. When string makes an angle of 35^o to vertical, tension in string = 6.1N, as illustrated.

Resultant force acting on stone in position shown:

Either

Vertical component: 6.1cos(35) = 4.99N

So, it has no resultant vertical force (since weight = 5.0N)

Other component: 6.1sin(35) = 3.5N horizontally

Or

A scale is shown. A triangle of correct shape is drawn. Resultant = 3.5 ± 0.2N Horizontal ± 3^o.

Question 4

{Detailed explanations for this question is available as Solution 1024 at Physics 9702 Doubts | Help Page 214 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-214.html}

 Question 5
{Detailed explanations for this question is available as Solution 975 at Physics 9702 Doubts | Help Page 201 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-201.html}

Question 6

2 horizontal metal plates X and Y are at distance 0.75cm apart. Positively charged particle of mass 9.6x10^-15kg situated in vacuum between plates, as illustrated. Potential  difference between plates adjusted until particle remains stationary.

(a)

Which plate, X or Y, is positively charged:

The electric force must be upwards (on the positive charge). So, plate Y is positive.

(b)

Potential difference required for particle to be stationary between plates is found to be 630V.

(i)

Electric field strength between plates:

E = V / d = 630 / (0.75x10^-2) = 8.4x10⁴NC^-1

(ii)

Charge on particle:

(Electric force = Gravitational force)

qE = mg

(g used to 1 dp since other values also given to 1dp)

q = mg / E = [(9.6x10^-15) x 9.8] / (8.4x10⁴) = 1.12x10^-18C

 

Question 7

{Detailed explanations for this question is available as Solution 911 at Physics 9702 Doubts | Help Page 186 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-186.html}

Question 8

{Detailed explanations for this question is available as Solution 1087 at Physics 9702 Doubts | Help Page 230 - http://physics-ref.blogspot.com/2015/12/physics-9702-doubts-help-page-230.html}