In head on collision of two bodies, derive the expressions for the velocities of the bodies in terms of their masses and velocities before collision.
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in head on collision velocity of first body is directed just opposite to velocity of 2nd body. or you can say that, when two bodies move in just opposite direction , and finally they collide . this type of collision is known as head on collision.
Let a body of mass m1, moves with speed v1 in east direction . another body of mass m2 moves with speed v2 in west direction. after interval of time, bodies collide at point.
from law of conservation of linear momentum, we know, if there is no any external force acts on system of bodies , linear momentum remains constant.
e.g., Initial linear momentum = final linear momentum
m1v1 - m2v2 = Lf
if we assume after collision , velocity of first body is v3 and velocity of 2nd body is v4
then, m1v1 - m2v2 = m1v3 + m2v4 ........(1)
[ here you should include sign of velocity ]
from work energy theorem,
initial kinetic energy = final kinetic energy
1/2m1v1² + 1/2m2v2² = 1/2m1v3² + 1/2mv4² .....(2)
here we have two unknown and two equations we can get v3 and v4 easily to solve them.
Let a body of mass m1, moves with speed v1 in east direction . another body of mass m2 moves with speed v2 in west direction. after interval of time, bodies collide at point.
from law of conservation of linear momentum, we know, if there is no any external force acts on system of bodies , linear momentum remains constant.
e.g., Initial linear momentum = final linear momentum
m1v1 - m2v2 = Lf
if we assume after collision , velocity of first body is v3 and velocity of 2nd body is v4
then, m1v1 - m2v2 = m1v3 + m2v4 ........(1)
[ here you should include sign of velocity ]
from work energy theorem,
initial kinetic energy = final kinetic energy
1/2m1v1² + 1/2m2v2² = 1/2m1v3² + 1/2mv4² .....(2)
here we have two unknown and two equations we can get v3 and v4 easily to solve them.
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