Two balls, of masses m A = 36 g and m B = 76 g are suspended as shown in Figure

Question

Two balls, of masses mA = 36 g and mB = 76 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60

Two balls, of masses mA = 36 g and mB = 76 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60 degree angle with the vertical and released. What is the velocity of the lighter ball before impact? (Take the right to be positive.) What is the velocity of each ball after the elastic collision? What will be the maximum height of each ball (above the collision point) after the elastic collision?

Explanation / Answer

a) Decrease in PE = Increase in KE

Ma* g* l(1-cos 60) = Ma*v*v/2

v = 1.7155 m/s

b) Conserving momentum :

0.036 * 1.7155 + 0 = 0.036 * Va + 0.076 * Vb

Conserving Energy :

0.036 * 1.7155*1.7155 / 2 + 0 = 0.036 * Va*Va/2 + 0.076 * Vb*Vb/2

Solving, Vb = 1.1 m/s ; Va = -0.6 m/s

c) using formula, mgh = m*v*v/2

Ha = 0.0183 m (Left)

Hb = 0.0617 m (Right)

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