Question

State and prove with the help of a neat labeled diagrams the law of conservation of
momentum

Answer 1

Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t.

A=m1(v1−u1) (change in momentum of particle A)

B=m2(v2−u2) (change in momentum of particle B)

FBA=−FAB (from third law of motion)

FBA=m2∗a2=m2(v2−u2)t FAB=m1∗a1=m1(v1−u1)t m2(v2−u2)t=−m1(v1−u1)t m1u1+m2u2=m1v1+m2v2

Therefore, above is the equation of law of conservation of momentum where m1u1+m2u2 is the representation of total momentum of particles A and B before the collision and m1v1+m2v2 is the representation of total momentum of particles A and B after collision..

Answer 2

Solution :

According to the law of conservation of momentum When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting. This law can be stated as Momentum is never created nor destroyed.

Proof :

Suppose 2 objects A and B each of mass $\sf\: m_{1}\: and m_{2}$ are moving initially with velocities $\sf\: u_{1}\: and u_{2}$, strike each after time t and starts moving with velocities $\sf\: v_{1}\: and v_{2}$

We know that,

★ Momentum = Mass × Velocity

Therefore,

Initial momentum of A = $\sf\: m_{1}u_{1}$

Initial momentum of B = $\sf\: m_{2}u_{2}$

Final momentum of A = $\sf\: m_{1}v_{1}$

Final momentum of B = $\sf\: m_{2}v_{2}$

★ Rate of change of momentum = Change in momentum/Time taken

Therefore,

$\sf \: F_{AB} = \dfrac{(m_1v_1 - m_1u_1)}{t}$

$\sf \: F_{AB} =m _1 \dfrac{(v_1 -u_1)}{t} ...i)$

Also, rate of change of momentum in B during collision

$\sf \: F_{BA} = \dfrac{(m_2v_2 - m_2u_2)}{t}$

$\sf \: F_{BA} =m _1 \dfrac{(v_2 -u_2)}{t} ...ii)$

From Newton's 3rd law or motion, we've :

$\sf \: F_{AB} = - F_{BA}$

$\longrightarrow \sf \:m _1 \dfrac{(v_1 -u_1)}{t} = - m _2 \dfrac{(v_2 -u_2)}{t}$

$\longrightarrow \sf \:m _1 v_1 - m_1u_1 = m _2v_2 + m_2u_2$

$\longrightarrow \sf \:m _1 u_1 + m_2u_2 = m _1v_1 + m_2v_2$

Thus, Total initial momentum = Total final momentum.