Particle 1 has a velocity v1 = 2 m/s and a mass m1 = 2 kg . this particle collides with particle 2 of mass m2 = 6 kg , which is at rest, and then the two particles stick together. the speed of the conglomerate particle is

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09-06-2018
Chemistry

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Particle 1 has a velocity v1 = 2 m/s and a mass m1 = 2 kg . this particle collides with particle 2 of mass m2 = 6 kg , which is at rest, and then the two particles stick together. the speed of the conglomerate particle is

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blahblahmali blahblahmali · Answer 1 · 2018-06-18T13:38:01+02:00

This follows the law of conservation of momentum. Momentum is the product of mass and velocity of object.
Momentum = mass(m) x velocity(v)
law of conservation of momentum means that the total momentum of system before the collision of 2 objects is equal to the total momentum after the collision
Before the collision total momentum
= m1v1 + m2v2
m1 = 2 kg
v1 = 2 m/s
m2 = 6 kg
v2 = 0 m/s
substituting the values in the equation
total momentum before = (2 kg x 2 m/s) + (6 kg x 0 m/s)
total momentum = 4 kgm/s
after the collision the 2 objects stick together and have a common velocity
total momentum after the collision = (6 kg + 2 kg)x V = 8V
V = speed of the conglomerate particle

since total momentum before is equal to total momentum after
8V = 4
V = 2 m/s
speed of conglomerate particle is 2 m/s