Newton’s Laws of Motion: Overview, Questions, Preparation

Newton's Law of Motion exhibits a relationship between the body and the forces acting on it. This force causes a change in the motion. Sir Issac Newton laid the foundation of mechanics in terms of three laws of motion.

Newton's Laws of Motion

Every body continues to be in its state of rest or uniform motion in a straight line unless compelled by some external force to act otherwise. So, if the net external force on a body is zero, its acceleration is zero.

Example

Consider a book at rest on a horizontal surface. It is subjected to two external forces, the force due to gravity (weight W) acting downward and upward force (R) on the book by the table. If the force on the object is equal and opposite, they are said to be balanced. Since the book is at rest, the net force is zero, which implies that force R must be equal and opposite to W.

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

Example

If two stones, light and heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The greater the speed, the greater the opposing force needed. Thus the mass and speed of the body is an important parameter that determines the effect of force on its motion. 

Under the action of force F for time interval Δt, the velocity of the body of mass m changes from v to v+Δv. So initial momentum p=mv changes to Δp =mΔv.

To every action, there is always an equal and opposite reaction.

Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies A and B. According to the third law, FAB = – FBA

Example

Newton's Laws of Motion for Class 11

Laws of Motion is the most important chapter. Questions will be on Newton's laws of motion, its examples, applications and problems based on the force, momentum and acceleration. The weightage of the topic is 10 marks.

1. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop?

Solution: F= -50 N(retardation force), m = 20 kg

a = F/m = -50/20 = -2.5ms⁻²

v = u +at ; u = 15, v =0, a= -2.5

t = -15/-2.5 =6s

2. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0ms⁻². Calculate the force of the blast.

Solution: M= 20000 kg = 2×10⁴ kg, a = 5

The thrust should overcome the force of gravity besides giving it an upward acceleration.

 g= 9.8

F = m(a+g) = 2×10⁴(5+9.8) =2.96 ×10⁵ N

3. Anmol, a 55 kg sprinter, starts with an acceleration of 3.5 m/s². What will be the net external force?

Solution: F = 55× 3.5 = 192.5N

 

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