An air track glider of mass m1 = 0.200kg moving at 0.750m/s to the right collides with a glider of mass m2 = 0.400kg at rest. if m1 rebounds and moves to the left with a speed of 0.250m/s, what is the speed and direction of m2 after the collision. direction is indicated by the sign of the velocity.

lets assume the right direction as positive direction of motion and left direction as negative direction of motion

m₁ = 0.200 kg

v₁ = velocity of m₁ before collision= 0.750 m/s

v'₁ = velocity of m₁ after collision= - 0.250 m/s

m₂ = 0.200 kg

v₂ = velocity of m₂ before collision= 0 m/s

v'₂ = velocity of m₂ after collision= ?

using conservation of momentum

m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂

(0.200) (0.750) + (0.400) (0) = (0.200) (- 0.250) + (0.400) v'₂

v'₂ = 0.500 m/s

so the glider of mass m₂ moves right at speed of 0.500 m/s

i believe the answer is electric potential.

explanation: