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