THIS IS DYNAMYCS QUESTION. ANSWER MUST BE EXACTLY 4 S.F PLEASEEEEE!!!!! Ball A,

Question

THIS IS DYNAMYCS QUESTION. ANSWER MUST BE EXACTLY 4 S.F PLEASEEEEE!!!!!

Ball A, of mass A-2.5kg, is attached to a taut cord of length L = 0.99m and released from an angle , = 360 It swings down along a circular arc until it has a direct, central impact with a stationary suspended ball B, of mass m 2.1kg The coefficient of restitution for the impact between the two balls is e 0.1 9-= 9.8m/s 2. Hint: The relationship between angle and speed at the apex can be determined from the work-energy principle. 0 a) To what angle ( /) will B swing to after the impact before it comes rest?

Explanation / Answer

Applying energy cosnervation to find speed of A at bottom (Just before hitting B)

m g L (1 - cos36) + 0 = 0 + m v^2 / 2

v= sqrt(2 x 9.81 x 0.99 (1 - cos36))

v = 1.93 m/s

Applying momentum conservation,

2.5 x 1.93 = 2.1 vB - 2.5 vA

2.1 vB - 2.5vA = 4.815 ..... . (i)

e = velocity of separation / velocity of approach

0.1 = vA + vB / 1.93

vA + vB = 0.193 ..... (ii)

Solving vA = - 0.96 m/s

vB = 2119/1840 = 1.16 m/s

Applyng energy conservation again on B:

1.16^2 /2 = 9.8 x 0.99 (1 - cos(theta))

cos(theta) = 0.932

theta = 21.3 deg .......Ans

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