Question

define momentum. state and prove law of conversation linear momentum.

Answer 1

Conservation of linear momentum states that if no external force acts on a body or system of a bodies,their linear momentum remains constant.
Consider two masses m1 and m2 moving with velocities v1 and v2 on a frictionless horizontal surface on the same line.
F12 bar and F21 bar are action reaction forces acting on the masses. After collision they are moving with velocity v1 bar and v2 bar. As there is no external force acting on this masses, their momentum must be conserved.
By Newton's second law, we know that force acts on object 1,
i.e. F12 bar = rate of change of momentum of m1
F12 bar = m1v1 bar - m1v1 bar ÷ t ( first equation )
F21 bar = rate of change of momentum of m2
F21 bar = m2v2 bar - m2v2 bar ÷ t (second equation )
But as per newtons third law ,
F12 bar = - F21 bar
m1v1 bar - m1v1 bar ÷ t = - m2v2 bar - m2v2 bar ÷ t so here t and t gets cancel out
m1v1 bar + m2v2 bar = m1v1 bar + m2v2 bar
So this is what conservation of linear momentum. :)

Answer 2

Conservation of linear momentum states that if no external force acts on a body or system of a bodies,their linear momentum remains constant.
Consider two masses m1 and m2 moving with velocities v1 and v2 on a frictionless horizontal surface on the same line.
F12 bar and F21 bar are action reaction forces acting on the masses. After collision they are moving with velocity v1 bar and v2 bar. As there is no external force acting on this masses, their momentum must be conserved.
By Newton's second law, we know that force acts on object 1,
i.e. F12 bar = rate of change of momentum of m1
F12 bar = m1v1 bar - m1v1 bar ÷ t ( first equation )
F21 bar = rate of change of momentum of m2
F21 bar = m2v2 bar - m2v2 bar ÷ t (second equation )
But as per newtons third law ,
F12 bar = - F21 bar
m1v1 bar - m1v1 bar ÷ t = - m2v2 bar - m2v2 bar ÷ t so here t and t gets cancel out
m1v1 bar + m2v2 bar = m1v1 bar + m2v2 bar
So this is what conservation of linear momentum.