Question

derive the equation of law of conservation of momentum​

Answer 1

$\underline{\huge\text{Question:}}$

Derive the equation of law of conservation of momentum .

$\underline{\huge\text{Answer:}}$

The answer to this question is :-

Momentum before collision = Momentum after collision .

$\underline{\huge\text{Step-By-Step-Explanation}}$

Suppose two bodies , a bus and a car , are moving in the same direction , with different velocities .

Let , the mass of the bus be $m_1$ and its velocity be $u_1$ so that its initial momentum is $m_1u_1$ .

Let , the mass of the car be $m_2$ and its velocity be $u_2$ so that its initial momentum is $m_2u_2$ .

Thus , the total momentum of the bus and the car before collision is $m_1u_1 + m_2u2$ .

Suppose the bus and the car collide for time , $t$ . Due to collision , the velocities of the bus and the car will change .

Let , the velocity of the bus after collision be $v_1$ , and the velocity of the car after collision be $v_2$ . So the momentum of the bus after collision will be $m_1v_1$ , and the momentum of the car after the collision wil be $m_2v_2$ .

The total momentum of the bus and the car will be $m_1v_1 + m_2v_2$ .

Let's assume that during collision , a force $F_1$ is exerted on the car by the truck , and in return , the car exerts a force $F_2$ on the truck .

Acceleration of the car $= \frac{(v_2 - u_2)}{t}$

So , the force exerted on the car

$= F_1 = m_2 × \frac{(v_2 - u_2)}{t}$

Acceleration of the bus = $= \frac{(v_1 - u_1)}{t}$

So , the force exerted on the bus

$= F_2 = m_1 × \frac{(v_1 - u_1)}{t}$

According to the third law of motion ,

$\textbf{Action = - Reaction}$

So ,

$F_1=F_2$

$=>\:\frac{m_2(v_2 - u_2)}{t} =-\frac{m_1(v_1 - u_1)}{t}$

$=>\:{m_2(v_2 - u_2)} =-{m_1(v_1 - u_1)}$

$=>\:m_2v_2 + m_1v_1 = m_2u_2 + m_1u_1$

$=>\:m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Therefore ,

$\textbf{Total momentum before collision = Total momentum after collision}$

#Hence - Proved !