Newton’s law Questions And Answers (part 1)

Newton’s law Questions

1. (a) State Newton’s first law of motion  

(b) Distinguish between elastic collision and inelastic collision  

(c) A minibus of mass 2000kg traveling at a constant velocity of 36km/h collides with a stationary

car of mass 1000kg. The impact takes 2 seconds before the two move together at a constant

velocity for 20 seconds. Calculate:

(i) The common velocity

(ii) The distance moved after impact

(iii) The impulse force  

(iv) The change in kinetic energy

2.  State Newton’s second law of motion  

3.  State the law of inertia

4.  A footballer kicks a ball of 600g initially at rest using a force of 900N. If the foot was in

 contact with the ball for 0.1sec. What was the take off speed of the ball?

5.  State Newton’s third law of motion  

6.  (a) State Newton’s second law of motion  

(b) The figure below shows two mini buses A and B at a speed of 40m/s and 20m/s

respectively moving in opposite directions. They collided head on

 Determine the common speed of the vehicles if they stuck to each other

Newton’s law Answers

1.  (a) Newton’s first law:

– A body remains in its state of rest or uniform motion in a straight line unless acted upon by an external force  *TEZ*

(b) Elastic collision is one in which both kinetic energy and momentum are conserved, while

inelastic collision is one in which momentum is conserved, but kinetic energy is not  (2mks)

(c) Momentum before collision = momentum after collision  1mk)

(2000 x 10) + (1000 x 0) = (2000 + 1000)V

3000V = 20000

V= 6²/₃m/s

(ii) d = Velocity x time *TEZ*

    = 20 x 20 = 400m = 133¹/₃m  

3 3

(iii) Impulse = change in momentum

= 2000 (10-6²/₃m/s) – for the minibus

 Or= 1000 (6²/₃m/s-0) – for the car

= 6667NS

Impulse of force = Impulse

Time

  = 6667 = 3333.5N

  2

(iv) K.E before collision = ½ x 2000 x 10² = 100,000J

  K.E after collision = ½ x 3000 x (6²/₃)² = 66,666.7J

Change in K.E = (100,000 – 66666.7) J

= 33,333.3J  

2.  The rate of change of momentum is directly proportional to the external force acting on a

body it is in the direction of force

3.  A body continues in its initial state rest or uniform motion unless compelled by an external

force to make it behave differently.

4.   F = Ma

900 = 600a

100 a = 9000 = 900ms^{-2  (1mk)}

10

but a = v-u

t

at = v-u

 (900 x 0.1) = v = 90ms^{-1  (1mk)}

5.  For every action there is an equal and opposite reaction 1

6.   (a) The rate of change of momentum of a body is directly proportional to the resultant external

force producing the change and takes place in e direction of force.

(b) m₁u₁ – m₂u ₂ = (m₁ + m₂)

V = m₁u ₁ – m₂u₂

M₁ + m₂

= (2500 x 40) – (3500 x 20)

2500 + 3500

= 30000

6000 = 5ms^-1