Michah (mass 40kg) is rollerskating east at 6ms when he collides with Joey (mass 60kg) who is rollerskating west at 5ms. As they collide, they quickly push off each other in such a way that Michah travels west at 3ms. What is Joey’s speed once they push off each other?
If needed, g=10ms2
To find Joey's speed once they push off each other, we need to apply the principles of conservation of momentum.
The momentum before the collision and the momentum after the collision should be equal, as no external forces act on the system. Mathematically, this can be expressed as:
Initial momentum = Final momentum
The momentum of an object is given by the product of its mass and velocity. Therefore, we can write the equation for momentum as:
(Michah's mass * Michah's velocity) + (Joey's mass * Joey's velocity) = (Michah's mass * Michah's final velocity) + (Joey's mass * Joey's final velocity)
Plugging in the given values:
(40 kg * 6 m/s) + (60 kg * (-5 m/s)) = (40 kg * (-3 m/s)) + (60 kg * Joey's final velocity)
240 kg∙m/s - 300 kg∙m/s = -120 kg∙m/s + 60 kg∙Joey's final velocity
Simplifying the equation:
-60 kg∙m/s = 60 kg∙Joey's final velocity
Dividing both sides by 60 kg:
Joey's final velocity = -1 m/s
Since velocity is a vector quantity and the negative value indicates the opposite direction, Joey's speed once they push off each other is 1 m/s to the east.