Respuesta :
Answer:
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
