Part(B) A baseball of mass m_1 = 0.25 kg is thrown at another ball hanging from

Question

Part(B) A baseball of mass m_1 = 0.25 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.05 m. The second ball m_2 = 0.61 kg is initially at rest while the baseball has an initial horizontal velocity of V_1 = 2.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity.) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d. What is the angle that the string makes with the vertical at the highest point of travel in degrees?

Explanation / Answer

the speed of the m2 after collision is V2 = m1*v1/m2

the height to which m2 will reach after collision is h = v2^2/(2*g) = m1^2*v1^2/(2*m2^2*g)

then required distmce is d = L-m1^2*v1^2/(2*m2^2*g) = L- [(m1*v1)^2/(2*m2^2*g)]

h = m1^2*v1^2/(2*m2^2*g)

h = (0.25*0.25*2.5*2.5)/(2*0.61*0.61*9.81) =0.0535 m

B) cos(theta) = (L-h)/L = (1.05-0.535)/1.05 = 60.6 degrees

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