Marking Scheme Physics Paper 2 Section B (Example)

Physics module ( Answers - Forces in Equilibrium)

Revison Questions Answers

1. a) at maximum height, velocity is instantaneously zero. As the ball decelerates at 10   ms^-2, velocity reduces at rate 10ms^-1 . Hence, its velocity is zero after 3s.  

    

   b)  v = u² + 2as, v = 0 ms^-1, u = 30 ms^-1, a = -10 ms^-2 . Hence, s= 45m.

  

   c) it took 3 s to reach maximum height, 3 s to return its original position. Therefore, the total time taken is 6 s.

2. a) rate of increase in velocity = rate of change of velocity = acceleration (free fall)

       hence, rate of increase in velocity = 9.8 ms^-2
    b) v = u + at, v = 39.2 ms^-1 

    c) s = ut + 1/2 at²  Therefore, s = 240.1 m

3. inelastic collision, u = 202 ms^-1 

4.  inelastic collision, u = -58 ms^-1 = -208.8 kmh^-1

5. a) deceleration, a = 4 ms^-2 or a = - 4 ms^-2
    b) braking force, F = ma = 6000 N or F = - 6000 N 
   c) distance moved during the car, s =  50 m
  
6. a) impulse = -6.6 Ns
   b) impulsive force = -330 N


7. a) 2500 kg ms^-1 
   b) 2500 kg ms^-1
    c) Principle Conservation of Momentum
    d) inelastic collision, v = 1 ms^-1

 8. a) explosion, v = 0.72 ms^-1
      b) inelastic collision, mass wooden block = 0.348 kg.


2.9 Analysing Forces In Equilibrium

9.  Normal reaction force,R = 3444.72 N

     Frictional force, F          = 289.25 N

10. a) Weight = 500 N

     b) Tension of each rope = 32.3 N

11. a) F = 40 N

     b) a = 5 ms^-2

     c) Tension = 25 N

12.  Friction, F_R   = mg sin Ѳ                               

                           = 5 x 10 x sin 15°                                     

                           = 12.9 N                                                    

        Normal, F_N  = mg cos Ѳ
                         = 5 x 10 x cos 15°
                         = 48.3 N
 

13.  Resolving the force, F and W (= mg) parallel to the plane.

(The vertical component does not play a role because when the block is in motion, it will only move along the plane.)

F cos Ѳ    = mg sin Ѳ

F cos 30°  = 8 sin 30°

F         = 4.6 N

14.   Resolve the forces into their components.

a)       For horizontal component:                         

T sinѲ = 20 sin30°                                     

T sinѲ = 10……….(1)                               

For vertical component:

T cosѲ + 20 cos30° = 25

T cosѲ = 7.68………….(2)

(1) + (2):  Ѳ = 52.5°

b)      T sin52.5° = 10

          T = 12.6 N = tension rope B

15. Total force upwards = Total force downwards

                     T₁ sin50° = 60°

                               T₁ = 78.3N

     Total force to the left = total force to the right

                     T₁ cos50° = T₂

                            T₂     = 50.3N

16.   F = ma

       (50 - 40) sin 30° = 5.0 x a

                             a = 3 ms^-2

17.   Parallel to surface:                          

        F = mg sin 60° = 43.3 N                       

       
       Perpendicular to surface:
       F = mg cos 60° = 25 N

Therefore, F = ma, 43.3 = 5 x a
                                  a = 8.66 ms^-2

2.10 Work, Energy, Power and Efficiency


18. a) W = 3N x 2m = 6 Nm
      b) W = -3N x -2m = 6 Nm 


19. W = Fs cos Ѳ = 25 Nm


20. a) W = Fs = 40 J
     b) W = Fs = 80 J


21. a) W = Fs 
              = mgs 
              = 750 J
     b) Work = Energy
                  = 750 J

Revision Questions

Chapter 2.10 Work

 

Chapter 2.9 Forces in Equlibrium

Chapter 2.8 Understanding Gravity

Explain the acceleration due to gravity by solving the problems.

Define the gravitational field.

State gravitational field strength.

Define and differentiate mass and weight.

All objects are pulled towards the center of the earth by a force known as the earth’s gravitational force (force of gravity).

Any object dropped towards earth which falls under the influence of the earth’s gravitational force (without any influence of other external forces, such as air friction) is said to be going through a free fall.

In reality, free fall only occur in vacuum space.

An object undergoing free fall will fall at the rate of gravitational acceleration which is at a constant of 9.81 ms-2 at sea level.

The gravitational acceleration is not influenced by the size or mass of the object.

Weight is the force of gravity which is exerted on the Earth.

Since weight, W is the gravitational force acting on the object of mass,m that makes it falls with an acceleration, g . Therefore, by using the corresponding terms of Newton’s Second Law,

F = ma or W= mg

Comparison Between Weight and Mass

weight, W
= force of gravity on the object
= varies with the magnitude of gravitational field strength,g
= vector quantity
= derived quantity
= Newton (N)

mass, m
= amount of matter in the object
= constant everywhere
= scalar quantity
= based quantity
= kilogram (kg)



SUMMARY

1. A gravitational force is force that pulls objects towards each other.

2.When an object falls under the influence of gravitational force only, the object is in a state of free fall.

3. The acceleration of an object experiencing a free fall is known as gravitational acceleration, g and the unit is ms-2.

4. Gravitational field strength, g is defined as the magnitude of the gravitational force acting on a unit mass of an object in the field.

g=F/m (Nkg-1)

5. The gravitational acceleration has a constant value of 9.8 ms-2 near the surface of the earth and does not depend on the mass of the object.

6. The gravitational field strength also has a value of 9.8 Nkg-1 near the surface of the earth.

7. The weight of an object is defined as the gravitational force acting on the object that is, the product of mass and acceleration due to gravity.