(13%) Problem 7: A baseball of mass m1 = 0.47 kg is thrown at another ball hangi
ID: 1460006 • Letter: #
Question
(13%) Problem 7: A baseball of mass m1 = 0.47 kg is thrown at another ball hanging from the ceiling by a length of string L = 15 m. The second ball m2 = 0.62 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 35 m/s. After the collision the first baseball falls straight down (no horizontal velocity.) Part (a) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d. Part (b) What is the angle that the string makes with the vertical at the highest point of travel in degrees?Explanation / Answer
1.
using momentum conservation for m1 and m2 collision,
m1v1 + m2 x 0 = m1 x 0 + m2v
0.47 x 3.5 = 0.62v
v = m1v1 /m2 = 2.65 m/s
using energy conservation to find height m2 rises above,
mv^2 /2 = mgh
h = (m1v1/m2)^2 / 2g = (m1v1^2 )/ 2 m2^2 g = 2.65^2 / (2 x 9.81) = 0.359 m
distance form ceiling d = L - h
d = L - (m1v1^2 )/ 2 m2^2 g
Last one
2) h = L ( 1 - cos@)
1 - cos@ = 0.359 / 1.5 = 0.239
cos@ =0.761
@ = 40.46 deg
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