Posted by Amber on Saturday, April 7, 2012 at 8:40pm.
A 1.3 kg steel ball and 8 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O.
This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 1.3 kg block sitting
at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of friction between the block and shelf to be 0.89.
The acceleration of gravity is 9.81 m/s2. What is the velocity of the block just after impact?
Answer in units of m/s

Physics. PLEASE HELP!!  Joe, Saturday, April 7, 2012 at 9:07pm
6 m/s

Physics. PLEASE HELP!!  Elena, Sunday, April 8, 2012 at 11:24am
m1 = 1.3 kg, m2 = 1.3 kg k = 0.89, g = 9.81 m/s2
M1•g•h = m1•v^2/2
v=sqrt(2•g•h) = sqrt(2•9.81•8) = 12.53 m/s.
Law of conservation of linear momentum m1•v1 + 0 = m1•u1 +m2•v2.
Since m1=m2, u2 =12.53 m/s.
Given coefficient of friction is necessary for the second part of this problem which is usually the calculation of the distance of the block motion.
a = k•g,
s = v^2/2a = v^2/2 •k•g =(12.53)^2/2•0.89•9.81 = 9 m
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