ball hanging (8%) Problem 7: A baseball of mass m,-0.44 kg is thrown at another
ID: 1790587 • Letter: B
Question
ball hanging (8%) Problem 7: A baseball of mass m,-0.44 kg is thrown at another fromthe ceiling by a length of string L = 1.7 m. The second bal1m2 = 0.87 kg is initially at rest while the baseball has an initial horizontal velocity of v, = 3.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity.) Randomized Variables my 0.44 kg m2 = 0.87 kg L=1.7m V1 = 3.5 m/s 2ot 50% Part (a) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d. Gr -- 2g 2m29 2m29 At 2m2g Hint I give up! Submit Feedback: 30% deduction per feedback. Hints: 30% deduction per hint. Hints remaining: 2 50% Part (b) what is the angle that the string makes with the vertical at the highest point of travel in degrees?Explanation / Answer
momentum before collision = momentum after collision
m1*v1 + m2*v2 = m1*v1' + m2*v2'
v1 = 3.5 i m/s
v2 = 0
v1' = 0
v2' = ?
m1*v1 = m2*v2'
v2' = (m1/m2)*v1
KE of second ball after collision = PE at the poisition d
(1/2)*m2*v2'^2 = m2*g*(L-d)
(1/2)*((m1/m2)*v1)^2 = g*(L-d)
L - d = (m1*v1)^2/(2*m2^2*g)
d = L - (m1*v1)^2/(2*m2^2*g) <<<------ANSWER
costheta = d/L = ( L - (m1*v1)^2/(2*m2^2*g))/L = 0.16/1.7
theta = cos^-1(d/L)
theta = 84.5 degrees
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