State and prove law of conservation of momentum
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The law of momentum conservation can be stated as follows. For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
Cherry010:
I want to prove also
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Hey mate.......
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In an isolated system with only two objects, the change in momentum of one object must be equal and opposite to the change in momentum of the other object.
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
Hence proved
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In an isolated system with only two objects, the change in momentum of one object must be equal and opposite to the change in momentum of the other object.
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
Hence proved
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