➢ Effect of Force
Forces are mainly of two types:
(i) Balanced forces
(ii) Unbalanced forces
➢ Balanced Forces
➢ Unbalanced Forces
Unbalanced forces can do the following:
1. Newton’s First Law of Motion (Law of Inertia)
Illustration of Newton's First Law of Motion
- If any object is in the state of rest, then it will remain in rest until an external force is applied to change its state.
- Similarly, an object will remain in motion until an external force is applied over it to change its state.
- This means all objects resist changing their state. The state of any object can be changed by applying external forces only. (a) Newton’s First Law of Motion in Everyday Life
(b) Mass and Inertia
Inertia increases with an increase in mass and decreases with a decrease in mass.
2. Newton’s Second Law of Motion
Some explanations to understand the momentum:
(c) Unit of Momentum
p = kg × m/s ⇒ p = kgm/s
The momentum of an object which is in the state of rest:
Let, an object with mass ‘m’ is in the rest.
Since, the object is at rest, therefore, its velocity, v = 0
∵ Momentum = mass × velocity
⇒ p = m × 0 = 0
Thus, the momentum of an object in the rest, i.e. non-moving is equal to zero.
Newton's Second Law of MotionWe know that:
(e) Proof of Newton’s First Law of Motion from the Second Law
The first law states that if external force F = 0, then a moving body keeps moving with the same velocity, or a body at rest continues to be at rest.
⇒ F = 0
We know, F = m(v-u)/t
(b) Law of Conservation of Momentum
Example 1: A bullet of mass 20 g is fired horizontally with a velocity of 150 m/s from a pistol of mass 2 kg. Find the recoil velocity of the pistol.Solution:
► Mass (m1) of bullet = 20 g = 0.02 kg
► Mass (m2) of pistol = 2 kg
► v1 for bullet = 150
Initially, the bullet is inside the gun and it is not moving.
∵ Mass = m1+m2 = (0.02 + 2) kg = 2.02 kg, and u1 = 0
⇒ Initial momentum = 2.02 × 0 = 0 ....(i)
Let the velocity of the pistol be v2.
⇒ Final momentum = m1v1 + m2v2 = 0.02×150 + 2v2 ...(ii)
∵ Initial momentum = Final momentum
⇒ (0.02×150)/100 + 2v2 = 0 [From equations (i) and (ii)]
⇒ 3 + 2v2 = 0
⇒ 2v2 = −3
⇒ v2 = −1.5 m/s.
(−) ve sign indicates that gun recoils in direction opposite to that of the bullet.
Example 2: Two hockey players viz A of mass 50 kg is moving with a velocity of 4 m/s and another one B belonging to the opposite team with mass 60 kg is moving with 3 m/s, get entangled while chasing and fall. Find the velocity with which they fall down and in which direction?A Collision of two hockey players: (a) before the collision and (b) after the collisionSolution:
Given, mA = 50 kg, uA = 4 m/s, mB = 60 kg, uB = 3 m/s.
Initial momentum A = mAuA = 50 × 4 = 200 kgm/s,
Initial momentum B = mBuB = 60 × 3 = 180 kgm/s.
⇒ Total initial momentum = 200 + 180 = 380 kgm/s ....(i)
⇒ Final momentum = (mA + mB)v = (50 + 60)v = 110v ....(ii)
According to the law of conservation of momentum:
Initial momentum = Final momentum
⇒ 380 = 110v
⇒ v = 380/110 = 3.454 m/s.