Question

A glider of mass 0.155 kg is moving to the right on a frictionless, horizontal air track with a speed of 0.760 m/s . It has a head-on collision with a glider 0.294 kg that is moving to the left with a speed of 2.29 m/s . Suppose the collision is elastic. Find the magnitude of the final velocity of the 0.155 kg glider. Find the magnitude of the final velocity of the 0.294 kg glider.

Answer 1

m1 = 0.155 Kg
v1 = 0.760 m/s

m2 = 0.294 Kg
v2 = 2.29 m/s

Let the Final Velocities of glider be v1f and v2f .

Initial Momentum = m1*v1 - m2*v2
Final Momentum = m1*v1f + m2*v2f.

Momentum Conservation -
Initial Momentum = Final Momentum
m1*v1 - m2*v2 = m1*v1f + m2*v2f
0.155 * 0.760 - 0.294 * 2.29 =  0.155*v1f + 0.294*v2f
-0.555 =  0.155*v1f + 0.294*v2f ----------1


As the Collision is Elastic, Kinetic Energy will be conserved -
0.5 * m1v1^2 + 0.5 * m2v2^2 = 0.5 * m1v1f^2 + 0.5 * m2v2f^2
0.155 * 0.760^2 + 0.294 * 2.29^2 =  0.155*v1f^2 + 0.294*v2f^2
1.63 = 0.155*v1f^2 + 0.294*v2f^2 ----------2

Solvinf for vf1 & vf2 using Eq 1 & 2 -
vf1 = -3.23 m/s
vf2 = -0.18 m/s

-ve sign means They are moving in the Left Direction.

Magnitude of the final velocity of the 0.155 kg glider = 3.23 m/s
Magnitude of the final velocity of the 0.294 kg glider = 0.18 m/s