question 1 :Give experimental verification of Newton’s third law of motion.
question 2 :Derive relation for law of conservation of momentum.
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Answers
Answer:
Step-by-step explanation:
Q1. Newton's Third Law states that in an interaction between two bodies A and B, the force exerted on A by B is equal and opposite to the force exerted on B by A at all times t during the interaction
Q2. Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t. B=m_{2}(v_{2}-u_{2}) (change in momentum of particle B)
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- According to Newton's third law, for every action there is an equal but opposite reaction. So, since the air pushed backwards, there was also a reaction force pushing the balloon forward. The air in the balloon created the action force. The more air in the balloon, the greater the force.
- Momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
- DERIVING THE LAW OF CONSERVATION OF MOMENTUM-
Let,
mA = Mass of ball A
mB= Mass of ball Ba
uA= initial velocity of ball A
uB= initial velocity of ball B
vA= Velocity after collision of ball A
vB= Velocity after collision of ball B
Fab= Force exerted by A on B
Fba= Force exerted by B on A
Now,
Change in momentum of A= momentum of A after collision - momentum of A before collision
= mA vA - mA uA
Rate of change of momentum A= Change in momentum of A/ time taken
= mA vA - mA uA/t
Force exerted by B on A (Fba)=
Fba= mA vA - mA uA/t. [i]
In the same way,
Rate of change of momentum of B=
mV vB - mB uB/t
Force exerted by A on B (Fab)=
Fab= mB vB - mB uB/t. [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
Fab= -Fba [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]
Using [i] and [ii] , we have
mB vB - mB uB/t = - (mA vA - mA uA/t)
mB vB - mB uB= - mA vA + mA uA
Finally we get,
mB vB + mA vA = mB uB + mA uA
This is the derivation of conservation of linear momentum.
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