A0.140-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. the glider has a head-on collision with a 0.299-kg glider that is moving to the left with a speed of 2.28 m/s. find the final velocity (magnitude and direction) of each glider if the collision is elastic.

The final velocity of the first glider is 3.39553 m/s in the opposite direction

The final velocity of the second glider is 0.31553 m/s in the same direction

Explanation:

= Mass of first glider = 0.14 kg

= Mass of second glider = 0.299 kg

= Initial Velocity of first glider = 0.8 m/s

= Initial Velocity of second glider = -2.28 m/s

= Final Velocity of first glider

= Final Velocity of second glider

As momentum and Energy is conserved

From the two equations we get

The final velocity of the first glider is 3.39553 m/s in the opposite direction

The final velocity of the second glider is 0.31553 m/s in the same direction

v1 = 2.76 m/s and v2 = - 0.32 m/s

Explanation:

m1 = 0.140 kg

m2 = 0.299 kg

u1 = 0.80 m/s

u2 = - 2.28 m/s

Let the speed after collision is v1 and v2.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2

0.112 - 0.68 = 0.14 v1 + 0.299 v2

0.14 v1 + 0.299 v2 = - 0.568 ..... (1)

By the use of coefficient of restitution, the value of e = 1 for elastic collision

u2 - u1 = v1 - v2

- 2.28 - 0.8 = v1 - v2

v1 - v2 = 3.08

v1 = 3.08 + v2

Put in equation (1)

0.14 (3.08 + v2) + 0.299 v2 = - 0.568

0.43 + 0.44 v2 = - 0.568

v2 = - 0.32 m/s

and

v1 = 3.08 - 0.32 = 2.76 m/s

Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s

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