Question

.
Derive
the
equation for
Conservation of momenturn?​

Answer 1

Answer:

Consider 2 objects A and B of masses ma and mb .Let they are traveling with the initial velocities of ua and ub. Let ua is greater than ub. Let them collide and collision lasts of time t. Let the final velocities be va and vb. Let there is no external unbalanced force applied on them.

By 3rd law of motion  

FAB = -FBA (1)

By 2nd law of motion  

F = ma. (2)

By 1st equation of motion

a = v-u/t. (3)  

Using (2) and (3) in (1)

FAB = ma(va-ua/t). (4)

Similarly  

FBA = mb(vb-ub/t). (5)

Using (4) and (5) in (1)

ma(va-ua/t) = -(mb(vb-ub/t))

t will get cancelled

(mv)a - (mu)a = -(mv)b + (mu)b

maua + mbub = mava + mbvb

Explanation:

hope it will help

Answer 2

Answer:

Law of conservation of momentum states that

For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

The law of conservation of momentum is an important consequence of Newton’s third law of motion.

Derivation of Conservation of Momentum

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 the collision