Newton's Laws of Motion Questions and Answers - Physics Form 3 Topical Revision

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  1. Two masses of 3kg and 7kg are connected by a light string. The 3 kg mass rests on a smooth incline plane 300 to the horizontal. The 7 kg mass hangs freely from the frictionless pulley attached to the top of plane.
    1. Draw a diagram showing the bodies and identify the forces acting on the 3 kg mass.
    2. Calculate the acceleration of the masses.
  2. A rocket propelled upward with a constant thrust. Assuming friction due to air is negligible and the burning of the fuel is steady. Explain its motion.
  3. A 2 kg body slides down a smooth slope from a height of 5m. As it reaches the horizontal, it strikes another body of mass 3 kg which is at rest. Both bodies stick together. Calculate the velocity of the bodies after collision.
  4. A girl of mass 40 kg stands on a scale balance in a lift. The lift is accelerating upwards. At one instant the acceleration of the lift is 2 m/s. Calculate the reading on the scale at that instant.
  5. The diagram below shows a tall measuring cylinder containing a viscous liquid. A very small steel ball is released from rest at the surface of the liquid as shown.
    cylinder with viscou YVbz6
    Sketch the velocity- time graph for the motion of the ball from the time it is released to the time just before it reaches the bottom of the cylinder.
  6. A body of mass 5 kg is ejected vertically from the ground when a force of 600N acts on it for 0.1s. Calculate the velocity with which the body leaves the ground.
      1. A body is initially in motion. If no external force acts on the body, describe the subsequent motion.
      2. A car of mass 800 kg is initially moving at 25 m/s. Calculate the force needed to bring the car to the rest over a distance of 20 m.
    2. Two trolleys of masses 2 kg and 1.5 kg are traveling towards each other at 0.25 m/s and 0.40 m/s respectively. Two trolleys combine on collision.
      1. Calculate the velocity of the combined trolleys.
      2. In what direction do the trolleys move after collision?
    1. The diagram below shows a block of mass 5 kg sliding down from rest on a plane incline at an angle of 300 to the horizontal. A frictional force of 6N acts between the block and the plane.
      1. Copy the diagram and show the forces acting on the block.
      2. Calculate the resultant force on the block.
      3. Calculate the time taken by the block to cover the distance of 25cm.
    2. The table shows the value of the resultant force, F, and the time t for a bullet traveling inside the gun barrel after the trigger is pulled.
       Force, F (N)  360 340  300  240  170   110
       Time, (t) (milliseconds)  3  12  17  22

      1. Plot a graph of Force, f against time t.
      2. Determine from the graph.
        1. The time required for the bullet to travel the length of the barrel assuming that the force becomes zero just at the end of the barrel.
        2. The impulse of the force.
        3. Given that the bullet emerges from the muzzle of the gun with a velocity of 200 m/s, calculate the mass of the bullet.
  9. The diagram below shows two identical strings A and B attached to a large mass M. String A is attached to the ceiling. State the reason why string B cuts when its free and is suddenly pulled downward.
    netwons law q9
  10. The figure below shows a 2 kg block attached to 0.5 kg mass by a light inextensible string which passes over a pulley. The force of friction between the horizontal bench and block is 3N. The block is released from rest so that both masses move through a distance of 0.6m. Calculate the velocity of the string.
    netwons lawq10
  11. A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks to it. State with a reason what is observed about the motion of the trolley.
  12. The figure below shows a block of mass 30.0 kg being pulled up a slope by force P at a constant speed. The frictional force on the block is 20.0N
    netwons law q12
      1. On the same figure name and indicate other forces acting on the block.
      2. Determine the component of the weight acting on the trolley down the slope
      3. Determine the value of P.
    2. On reaching the top of the slope, the block is left to run freely down the slope.
      1. Which one of the forces previously acting on the block would then act in the opposite direction?
      2. Determine the acceleration of the block down slope.
      3. What is the effect of increasing the angle of slope on your answer in (ii) above?
  13. A high jumper usually lands on a thick soft mattress. Explain how the mattress helps in reducing the force of impact.
  14. A resultant force F acts on a body of mass m causing an acceleration a1 on the body. When the same force acts on a body of mass 2m, it causes an acceleration a2. Express a2 in terms of a1.
    newtons lawq15
    The figure above shows two trolleys of mass 2.0 kg and 1.5 kg traveling towards each other at 0.25 m/s and 0.4 m/s respectively. The trolleys combine on collision. Calculate the velocity of the combined trolley and show the direction in which they move after collision.
  16. Two identical stones A and B are released from the same height above the ground. B falls through air while A falls through water. Sketch the graphs of kinetic energy (KE) against time (t) for each stone. Label the graph appropriately.
  17. A trolley is moving at uniform speed along a track. A piece of plasticine is dropped on the trolley and sticks on it. Explain why the trolley slows down. (2 mks)
    1. State the Newtons law of motion. (1 mk)
    2. A wooden block resting on a horizontal bench is given an initial velocity, u, so that it slides on the bench surface for a distance, d, before coming to a stop. The values of d were measured and recorded for various values of initial velocity. The figure below shows the graph of u2 against d.
      newtons law graph 18
      1. Determine the slope, s, of the graph
      2. Given that u2 = 20kd, where k is a constant for the bench surface, determine the value of k from the graph.
      3. State how the value of k would be affected by change in the roughness of the bench surface.
    3. A car of mass 800kg starts from the rest and accelerates at 1.2ms-2. determine its momentum after it has moved 400m from the starting 
  19. A force of 6N acts on a 2kg trolley and accelerates at 2 m/s2. Calculate the retarding force acting on the trolley.
  20. A boulder is sliding down a slope, with a uniform acceleration of 3 ms-2; calculate its velocity after it has slid 10m down the slope.
  21. A man whose mass is 70 kg stands on a spring weighing machine. When the lift starts to ascend its acceleration is 2.45m/s2. What is the reading on the scale?
  22. A bullet of mass 22 g traveling at a velocity of 18 m/s penetrates a sand bag and is brought to rest in 0.6 seconds. Find:
    1. The depth of penetration in metres 
    2. The average retarding force of the sand
  23. A bullet of mass 10g traveling horizontally with a velocity of 300 m/s strikes a block of wood of mass 290g which rests on rough horizontal floor. After impact they move together and come to rest after traveling a distance of 15m. Calculate
    1. the common velocity of the bullet and the block.
    2. the acceleration of the bullet and the block.
    3. the coefficient of sliding friction between the block and the floor.
    1. A body of mass m initially at rest is acted on by a force F for a time t, as a result its velocity changes to a final value V. Use this information to show that the gain is kinetic energy E= ½ MV2
    2. Calculate the kinetic energy of a car of mass 1000 kg traveling at 36 km/h
  25. Under a driving force of 400N a car of mass 1250 kg has an acceleration of 2.5 m/s. Find the frictional force acting on the car.
  26. An apple of mass 100g falls a distance of 2.5m to the ground from a branch of a tree.
    1. Calculate the speed at which it hits the ground and the time taken for it to fall. (Ignore air resistance).
    2. Assuming the apple takes 100 milliseconds to come to rest. Calculate the average force experienced by the apple.
  27. A helicopter of mass 3000 kg rises vertically at a constant speed of 25 ms-1 if the acceleration due to gravity is 10 ms-2; determine the resultant force working on the helicopter


      newtons law ans1
      R - Reaction
      W - Weight
      T - Tension
    2. Net force= 70N – mgsinx= 70N- 30×sin 30
      F= ma
      55= (7 + 3) a
      a= 5.5m/s2
  2. The rocket will accelerate due to 2 reasons
    • The total mass of the rocket decreases as the fuel burns but thrust force is constant
    • The gravitational pull decreases with increase in distance from the centre of the earth so it will be less.
  3. P.E = K.E
    Mgh = ½ mv2 therefore V= √(2gh) = √(20 x 5) = 10m/s
    Momentum before collision = Momentum after collision
    M1U1 = (M1 + M2)20
    10 x 25 x 20
    V= 4m/s
  4. Total force downward = weight of the girl = 400N. Since the lift is moving upwards the upward force T, is greater than 400N therefore from F=ma
    (T − 400) = 40 x 2
    T = 480N = reading
    newtons law graphans 5
  6. Force upward = 600N, Force downward = 50N (wt).Therefore Net force = 550N upwards. 
    F= ma = m(v-u)/t = m(v)/t since u= 0
    550 = (50 x v)/0.1
    V = 11m/s
      1. The body continues in its uniform state of motion velocity
      2. U= 25 m/s, V=0, S= 20m a=?
        V2 = U2 + 2as
        0= 625 + 40a
        a= -15.625m/s2
        F= ma = 800x – 15.625
        = -12500N
        Therefore decelerating force= 12500N
    2. Since one is moving towards the other one is +ve while the other is –ve.
      Momentum before collision = momentum after collision
      1. M1u1 + m2u2 = (m1+m2) v
        (2x 0.25) – (1.5 x 0.4) = 3.5 x v
        0.5 – 0.6 = 3.5 v
        v= -0.02867m/s
        = -2.867 x10-2m/s
      2. The original direction of the 1.5kg mass
        Fr= Frictional force
      2. Force= mgsin 30 − 6
        = 5 x 10 x 0.5 − 6
        = 19N
      3. F= ma
        19=5a = a = 3.8m/s2
        U= 0 m/s
        S= 0.25m
        t =?
        s= ut + ½ at2
        0.25 = 0+ 1.9t2
        t2 = 0.3627s
        = 0.36s
      1. Plot the graph
      2. Let it touch y and x – axis
        1. Time taken by bullet = x intercept (30ms)
        2. Impuse = force x time = area bound between x – axis and the line of the graph (6Ns)
        3. V= 200m/s        m=?
          Impulse = Ft = Mv – Mu
          = 6Ns = M x 200-0 since u= 0
          Therefore m= 0.03kg
  9. M tends to remain in its state of uniform rest due to its inertia. This protects string A
  10. Propelling force= 0.5 x 10= 5N
    Opposing force= frictional force = 3 N
    Net Force = 5-3 = 2N
    Using F= ma
    2= (2+ 0.5)a
    a= 0.8m/s2
    s= 0.6m
    V2 = U2 + 2as
    = 0+ 2+ 0.8 + 0.6
    = 0.96
    Therefore V= 0.9798
    V= 0.098m/s
  11. Speed decrease. Since momentum before collision is equal to momentum after collision i.e M1V1 = M2V2 then increase in mass implies/cause a decrease in velocity.
      1. R=Reaction
      2. Mg sin 10 = 30 x 10 sin 10
        = 52.1 + 20
      3. Since the velocity is constant then the net force = 0 
        p= mgsinӨ + Fr
        p= 52.1 + 20
        = 72. 1N
      1. Frictional force
      2. F= ma
        (52.1 − 2.0) = 30a
        32.1 = 30a
        a = 1.07 m/s2
  13. The mattress increase time taken to land. Thus from 
    f= (Change in momentum)
               Time taken
    When time is more the force that will decelerate the jumper will be smaller (safer)
  14. Fr= frictional force
    F= ma
    F= 2ma
    1 = a1
    i.e. a2 = (½)a1 OR read the value of x- intercept i.e. when a=0 m=mo and substitute in the equation to get m. 
  15. Since one is moving towards the other one has +ve vel while the other is –ve
    Momentum before collision = momentum after collision
    1. m1u1 + m2u2 = (m1m2)v
      (2 x 0.25) − (1.5 x 0.4) = 3.5 x v 
      0.5 – 0.6 = 3.5 v
      V= 0.02867m/s
      = -2.867 x 10-2 m/s
    2. The original direction of the 1.5 kg mass

    newtons law ans16
  17. Speed decreases. Since momentum before collision is equal to momentum after collision i.e. M1V1 = M2V2 then increase in mass implies/cause a decrease in velocity.
    1. A body at rest or in motion at uniform velocity tends to stay in that state unless acted on by an unbalanced force.
      1. Slope s= 97.5 - 0 (m/s2)
                       16 - 0 
        20k = s = 6.09
      2. k = 6.17/20 = 0.304
      3. Increase in roughness increase k and vice versa
    3. Applying equation
      V2 – U2 = 2as
      V2 – 0= 2 x 1.2 x 400
      Momentum P = mv
      = 800 x √(2 x 1.2 x 400)
      = 24787.09
      = 24790 (table)
  19. 2N
  20. 7.75m/s
  21. 2415N
    1. 5.4m
    2. 0.66N
    1. 10m/s,
    2. -3.33m/s,
    3. 0.333
    2. 50,000J
  25. 875N
    1. 7.07 m/s, 0.71s,
    2. 0.071kgm/s (ii) 0.71N (iii) 0.71N
  27. 30000N

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