Posted by **Amber** on Saturday, April 7, 2012 at 8:40pm.

A 1.3 kg steel ball and 8 m cord of negligi-ble 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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