Define the law of conservation of momentum.
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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. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
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The law of conservation of momentum states that the total momentume of two bodies before collision is equal to the total momentum of the objects after collision.
So,
if Mass of objects are: m1 and m2
let their initial velocities be u1 and u2
The two bodies collide and there is a chsnge in velocities. Let their final velocities be v1 and v2
so,
Initial momentume of first object is m1u1
and initial momentume of 2nd object is m2u2
Now,
final momentum of 1st object is m1v1
and final momentume of 2nd object is m2v2
Rate of change of momentum of object 1:
m1v1-m1u1/t (action)
And rate of chanege of momentume of second object is m2v2-m2u2/t (reaction)
By newton's 3rd law of motion,
action = -(reaction)
m1v1-m1u1/t = -m2v2 + m2u2
(HERE t gets cancelled on both sides)
so, m1v1 + m2v2 = m1u1 + m2u2
Hence Law of conseevation of momentume is derived.
HOOE IT HELPED U AND IF YES, PLZ MARKS AS BRIANLIEST!!!
So,
if Mass of objects are: m1 and m2
let their initial velocities be u1 and u2
The two bodies collide and there is a chsnge in velocities. Let their final velocities be v1 and v2
so,
Initial momentume of first object is m1u1
and initial momentume of 2nd object is m2u2
Now,
final momentum of 1st object is m1v1
and final momentume of 2nd object is m2v2
Rate of change of momentum of object 1:
m1v1-m1u1/t (action)
And rate of chanege of momentume of second object is m2v2-m2u2/t (reaction)
By newton's 3rd law of motion,
action = -(reaction)
m1v1-m1u1/t = -m2v2 + m2u2
(HERE t gets cancelled on both sides)
so, m1v1 + m2v2 = m1u1 + m2u2
Hence Law of conseevation of momentume is derived.
HOOE IT HELPED U AND IF YES, PLZ MARKS AS BRIANLIEST!!!
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