Question

Explain Impulse momentum theorem in detail.

Answer 1

Solution:

According to Impulse momentum theorem impulse is equal to change in momentum.

$\sf{\implies I = \Delta P = P_{f}-P_{i}}$

Derivation:

$\sf{\implies F = \dfrac{\Delta P}{\Delta t} = \dfrac{dP}{dt}\;\;\;\;.......(1)}$

$\sf{\implies F.dt=dP\;\;\;\;.......(2)}$

If momentum is changed from "i" to "f", then

$\sf{\implies \int\limits_{t_{1}}^{t_{2}}F.dt\;\;=\;\;\int\limits_{P_{i}}^{P_{f}}dp}$

$\sf{\implies F. \int\limits_{t_{1}}^{t_{2}}dt\;\; = \;\; \bigg(P \bigg)_{P_{i}}^{P_{f}}}$

$\sf{\implies F\bigg(t \bigg)_{t_{1}}^{t_{2}}=P_{f}-P_{i}}$

$\sf{\implies F.(t_{2}-t_{1})=P_{f}-P_{i}}$

$\sf{\implies F.\Delta t = P_{f}-P_{i}}$

${\boxed{\boxed{\sf{\implies I = P_{f}-P_{i}}}}}$

So, Impulse is equal to rate of change of momentum.

Extra Info:

Impulse:

• It is the product of average force and small time interval.
• I = F × Δt.
• I = (Newton × sec.)

Answer 2

Answer:

Explanation:

Impulse momentum Theorem;

This theorem is basically derived from Newton's Second Law of motion.

Impulse unit is Newton second.

According to this theorem the change in momentum ∆p = impulse applied to that particular body.

It is equivalent to second law of motion which says that force is directly proportional to force applied.

Hope it helps.

Radhe Radhe