Question

What is principle of conservation of momentum?? Prove it.

Answer 1

f12=change in momentum in 2/t

f12=m2v2-m2u2/t

f21=change in momentum in 1/t

f21=m1v1-m1u1/t

f12=-f21

m2v2-m2u2/t=-m1v1-m1u1/t

t WILL GET CANCEL..

m2v2+m1v1=m2u2+m1u1

THIS LAW OF CONSERVATION OF MOMENTUM..

follow me and mark me as brainlist answer.....please...please...

Answer 2

Solution:

Principle of conservation of Momentum:

=> It states that for an isolated system in the absence of external force momentum remains conserve that is before the collision momentum is equal to after the collision of momentum.

=> $\sf{m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}}$

Derivation:

Let us consider a system in which two objects A and B have initial speed $\sf{u_{1}}$ and $\sf{u_{2}}$ with mass $\sf{m_{1}}$ and $\sf{m_{2}}$ collide slightly and after the collision their velocities will be $\sf{v_{1}}$ and $\sf{v_{2}}$.

According to 3rd law of motion every action have equal and opposite reaction. i.e

$\sf{\implies \overrightarrow{F_{AB}}=-\overrightarrow{F_{BA}}}$

$\sf{\implies \Bigg(\dfrac{P_{f}-P_{i}}{t}\Bigg)_{A}=-\Bigg(\dfrac{P_{f}-P_{i}}{t}\Bigg)_{B}}$

$\sf{\implies \Bigg(\dfrac{m_{1}v_{1}-m_{1}u_{1}}{t}\Bigg)_{A}=- \Bigg(\dfrac{m_{2}v_{2}-m_{2}u_{2}}{t}\Bigg)_{B}}$

$\sf{\implies m_{1}v_{1}-m_{1}u_{1}=-m_{2}v_{2}+m_{2}u_{2}}$

$\large{\boxed{\boxed{\sf{\implies m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}}}}}$

=> After collision = Before collision.