Question

Derive and show that the rate of change of momentum is the product of mass and acceleration.

Answer 1

Consider a Body of Mass m having an Initial Velocity u. The Initial Momentum, p, of this Body will be mu. Suppose a Force F acts on this Body for Time t and causes the Final Velocity to become v. The Final Momentum $p_2$ of this Body will be mv.

$\bold{The\:change\:in\:Momentum\:\alpha\:p_2 - p_1}\\\\\alpha\: mv -mu\\\\\alpha\: m (v - u)\\\\\\\bold{The\:Rate\:of\:change\:of\:Momentum\:\alpha\:\frac{m (v - u)}{t}} \\\\According\:to\:Newton's\:Third\:Law\:of\:Motion,\\\\F\:\alpha\:\frac{m (v - u)}{t} \\\\\bold F = \frac{km * (v - u)}{t} = \bold{kma}$

Here, a = (v - u)/ t is the Acceleration, which is the Rate of change of Velocity. The quantity, k is a Constant of Proportionality. The SI Units of Mass and Acceleration are kg & m/s^2 respectively. The Unit of Force is so chosen that the Value of the Constant, k becomes 1.

For this, one Unit of Force is Defined as the amount that produces an Acceleration of 1 m/s^2 in an Object of 1 kg mass, (i.e.),

$1\:Unit\:of\:Force = k * (1\:kg) * (1\:m/s^2)$

Thus, the Value of k becomes 1. From F = kma, we get -

$\bold{F = ma}$

Hence,

It is Proved that the Rate of Change of Momentum is the Product of Mass and Acceleration.