A 0.157kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.850m/s . It has a head-on collision with a 0.306kg glider that is moving to the left with a speed of 2.26m/s . Suppose the collision is elastic. Find the magnitude of the final velocity of the 0.157kg glider. Find the magnitude of the final velocity of the 0.306kg glider.

V1 = -3.260 m/s,  V2 = 1.303 m/s

Explanation:

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find:  Final Velocity of Left Glider is V1=? m/s  and  Velocity of right Glider is V2 =? m/s  (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg)  + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg)  + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149,  0.463

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length