Question

State Newton's third law of motion. Also give prove F=ma (k=1)​

Answer 1

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Q1. State Newton's Third Law of Motion.

⟹ If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

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Or

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Every action has its equal and opposite reaction.

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Q2. Prove that F = ma.

⟹ Let us consider an object of mass m, moving along a straight line with an initial velocity u. Let us say, after a certain time t, with a constant acceleration, the final velocity becomes v. Here we see that, the initial momentum

p1 = m × u

The final momentum

p2 = m × v

The change in momentum can be expressed as

p2 – p1 = (m × v) – (m × u)

p2 – p1 = m (v – u)

But we know that the rate of change of momentum with respect to time is proportional to the applied force.

The applied force

F ∝ [m (v – u)]/t

F ∝ m × a

as acceleration (a) = rate of change of velocity with respect to time.

Answer 2

Step-by-step explanation:

NEWTON'S THIRD LAW OF MOTION:-

For two different bodies, every action has an equal and opposite reaction.

Proof of F=ma

Let there be an object with mass m, moving in a straight line, with initial velocity u. After a certain time t, then the velocity becomes a final velocity v, then we get the initial momentum as:-

$p1 = m \times u$

And the final momentum as:-

$p2 = m \times v$

The change is momentum is expressed by taking both LHS and RHS like this:-

$p2 - p1 = (m \times v) - (m \times u) \\ p2 - p1 = m(v - u) \: \: \: (taking \: m \: as \: common)$

But the rate of change of momentum with respect to time is directly proportional to the force applied.

$f \infty \: \frac{m(v - u)}{t} </strong></p><p><strong>(</strong><strong>Please</strong><strong> </strong><strong>consider</strong><strong> </strong><strong>the</strong><strong> </strong><strong>infinity</strong><strong> </strong><strong>sign</strong><strong> </strong><strong>as</strong><strong> </strong><strong>the</strong><strong> </strong><strong>proportionality</strong><strong> </strong><strong>sign</strong><strong>)</strong></p><p><strong>As</strong><strong> </strong><strong>acceleration</strong><strong> </strong><strong>=</strong><strong> </strong><strong>Rate</strong><strong> </strong><strong>of</strong><strong> </strong><strong>change</strong><strong> </strong><strong>of</strong><strong> </strong><strong>velocity </strong><strong>with</strong><strong> </strong><strong>respect</strong><strong> </strong><strong>to</strong><strong> </strong><strong>time</strong><strong>,</strong></p><p><strong>[tex]f = kma$

Hence proved.