Question

derive the conservation of momentum
MAUA+MBUB=MAVA+MBVB

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

Explanation:

Last line: m(v)a-m(u)a =-(mv)b+(mu)b

maua+mbub = mava-mbvb.