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.

a) -3.3 or 0.88m/s b)-0.13m/s or -2.3 m/s

Explanation:

using conservation of momentum

where M₁ = mass of glider 1 and M₂ = mass of glider 2, U₁ = initial velocity of glider 1 and U₂ = initial velocity of glider 2, V₁ = final velocity of glider 1 and V₂ = final velocity of glider 2

using conservation law of momentum

M₁U₁ + M₂U₂ = M₁V₁ + M₂V₂

substituting the values into the equation

(0.157 × 0.85) + ( 0.306 ×-2.26) = 0.157V₁ + 0.306V₂

-0.558 = 0.157V₁ + 0.306V₂

also the sum of their kinetic energy before and after collision are the same since the collision is elastic

1/2M₁U₁² + 1/2M₂U₂² = 1/2 M₁V₁² + 1/2M₂V₂²

remove half from both sides and substitute the values into the resulting equation

(0.157 × 0.850²) + ( 0.306× -2.26²) = 0.157 V₁² + 0.306V₂²

0.113 + 1.56 =  0.157 V₁² + 0.306V₂²

-0.558 = 0.157V₁ + 0.306V₂

-0.558 - 0.157V₁ = 0.306V₂

divide both side by 0.306

-0.558/0.306 - 0.157V₁/0.306 = V₂

-1.82 - 0.513V₁ = V₂

substitute V₂ into equation 2

1.673 = 0.157 V₁² + 0.306V₂²

1.673 = 0.157 V₁² + 0.306 (-1.82 - 0.513V₁)²  

1.673 = 0.157V₁² + 0.306( 3.3124 + 0.93V₁ + 0.93V₁ + 0.26V₁²)

1.673 = 0.157V₁² + 1.01 + 0.57V₁ + 0.08V₁²

1.673 = 0.157V₁² + 0.08V₁²+ 0.57V₁  + 1.01

0 = 0.157V₁² + 0.08V₁²+ 0.57V₁ + 1.01 - 1.673

0 = 0.237V₁²+ 0.57V₁ - 0.663

using quadratic formula calculate V₁

-b±√(b² - 4ac) / 2a

-0.57 ± √ (0.57² - (4 × 0.237×-0.663)) / (2×0.237)

-0.571± √(0.3249 + 0.629) / 2× 0.237

-0.571 ± √0.977 / 0.474

-1.56/ 0.474 = -3.3 or 0.88m/s

b) substitute the values into equation 1

-1.82 - 0.513V₁= V₂

V₂ = -0.13m/s or -2.3 m/s

dang that balls went far