CBSE Class 9 Physics Force and Laws of Motion Notes Set 03

Newton's First Law and Inertia

A body continues in its state of rest or of uniform motion in a straight line unless an external force acts on it. This law gives the idea of force (that which changes the state of rest or motion) and of inertia.
1. Inertia is the inability of a body to change its state of rest or uniform motion by itself. It is measured by mass: the greater the mass, the greater the inertia.
2. Examples: a coin drops into a bottle when the card under it is flicked away; a passenger lurches forward when a moving bus stops suddenly and sideways when it takes a turn; a hammer head tightens on its handle when the handle is struck on the ground.
3. If the net force on a body is zero, its velocity does not change; a body moving at constant velocity on a frictionless table has no net force on it.

Newton's Second Law and Momentum

The rate of change of momentum of a body is proportional to the applied force and takes place in the direction of the force. Momentum is the product of mass and velocity, p = mv.
1. For constant mass, \( F = ma = \dfrac{m(v - u)}{t} \). The SI unit of force is the newton: 1 N = 1 kg m/s².
2. Impulse = force x time = change of momentum = m(v - u), in N s. A large force acting for a very short time is an impulsive force.
3. Weight is the force of gravity, W = mg. On a support accelerating downward with acceleration a, the apparent weight is m(g - a); it is zero in free fall.
4. A given force acting for the same time on bodies of different masses produces the same change of momentum but accelerations inversely proportional to the masses.

Newton's Third Law and Conservation of Momentum

To every action there is an equal and opposite reaction, and the two act on different bodies. Examples: a boy jumping ashore pushes the boat backward; a gun recoils when a bullet is fired; we move forward because the ground pushes us forward when we push it backward.
1. When two bodies interact with no external force, the total momentum is conserved: \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \).
2. Gun and bullet: \( m_1v_1 = -m_2v_2 \), so the gun recoils with velocity \( v_2 = -\dfrac{m_1v_1}{m_2} \), much smaller than the bullet's velocity because the gun is far heavier.
3. When two bodies stick together after colliding, \( m_1u_1 + m_2u_2 = (m_1 + m_2)v \).
4. A machine gun firing n bullets of mass m each second with speed v experiences an average recoil force nmv.

Worked Example

Example. A cannon ball of 2 kg leaves a cannon of mass 500 kg with a speed of 100 m/s. The cannon moves back 0.5 m before stopping. Find the resistance offered by the ground.
Recoil velocity: \( v_2 = \dfrac{2 \times 100}{500} = 0.4 \) m/s. Retardation: \( a = \dfrac{v^2}{2s} = \dfrac{0.16}{1} = 0.16 \) m/s². Resistance = \( 500 \times 0.16 = 80 \) N.