Question

Derive and prove that rate of change of momentum is force​

Answer 1

Let an object of mass 'm' is moving along a straight line with initial velocity 'u'. A constant force 'F' is applied in time 't' to accelerate it and its final velocity becomes 'v'.

Initial momentum, p

1

=mu

Final momentum, p

2

=mv

Change in momentum ∝p

2

−p

1

∝mv−mu

∝m(v−u)

The rate of change of momentum =

t

m(v−u)

Rate of change of momentum = force applied

Force=

t

m(v−u)

Force=k

t

m(v−u)

where k = proportionally constant

Force = ma where a = acceleration =

t

(v−u)

Explanation:

i hope it's helpful to you

Answer 2

Answer:

Let initial momentum ($p_i$) be mu

Let final momentum ($p_f$) be mv

According to 2nd law of motion

$\frac{p_f - p_i}{t} \propto \: f$

$\implies \: f \propto \frac{mv \: - mu}{t}$

$\implies \: f \propto \frac{m(v - u)}{t}$

$f \propto \: ma \: \: \: \: \: \: ( \frac{v - u}{t } = a )$

To remove the proportionality sign. We would add k as the proportionality constant

$f = kma \\ f = ma \:$

because by the definition of force k = 1