A block lies on a smooth plane tilted at an angle 30 deg to the horizontal. (a) determine the acceleration of the block as it slides down the plane. (b) if the block starts from rest 9.10 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline? ignore friction
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someone confence fir the derivation of momentum please share it.....This law states that the sum of the momenta before collison is equal to the sum of momenta after collision,provided no external unbalanced force acts on it.
Let two bodies have mass m1 and m2 moving with a initial velocity u1 and u2 respectively.After collision the first body attains a final vecity v1 and the second body attains the final velocity as v2.
The momentum of the first body and second body before collision is:-
P1=m1u1 P2=m2u2
The momentum of the first body and second body after collision is:-
P1 '=m1v1 P2 '=m2v2
force exerted by the first body on the second body is
F12=P1 '-P1/t
F12=m1v1-m1u1/t
similarly
force exerted by the second body on the first body is
F21=P2 '-P2/t
F21=m2v2-m2u2/t
According to newtons third law of motion, the force exerted by the first body on the second body should be equal and opposite.
Thus,
=>F12= -F21
=>m1v1-m1u1/t= -(m2v2-m2u2)/t
=>m1v1-m1u1= -m2v2+m2u2 (by cancelling "t"on both sides)
=>m1v1+m2v2=m1u1+m2u2
=>m1u1+m2u2=m1v1+m2v2
thus,
the total momentum of the two bodies before collision is equal to the total momentum of the two bodies after the collision.
Hence,the momentum of two bodies are conserved,provided no external force acts on it.
Let two bodies have mass m1 and m2 moving with a initial velocity u1 and u2 respectively.After collision the first body attains a final vecity v1 and the second body attains the final velocity as v2.
The momentum of the first body and second body before collision is:-
P1=m1u1 P2=m2u2
The momentum of the first body and second body after collision is:-
P1 '=m1v1 P2 '=m2v2
force exerted by the first body on the second body is
F12=P1 '-P1/t
F12=m1v1-m1u1/t
similarly
force exerted by the second body on the first body is
F21=P2 '-P2/t
F21=m2v2-m2u2/t
According to newtons third law of motion, the force exerted by the first body on the second body should be equal and opposite.
Thus,
=>F12= -F21
=>m1v1-m1u1/t= -(m2v2-m2u2)/t
=>m1v1-m1u1= -m2v2+m2u2 (by cancelling "t"on both sides)
=>m1v1+m2v2=m1u1+m2u2
=>m1u1+m2u2=m1v1+m2v2
thus,
the total momentum of the two bodies before collision is equal to the total momentum of the two bodies after the collision.
Hence,the momentum of two bodies are conserved,provided no external force acts on it.
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