What is the speed of the ball just before impact
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What is the speed of the ball just before impact

[From: ] [author: ] [Date: 12-12-08] [Hit: ]
60 kg ball that is at rest, as shown.conservation of mechanical energy, find the speed of the 1.impact.So,......
Here's the disgram; http://tinypic.com/r/167plj5/6

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings
downward and strikes a 4.60 kg ball that is at rest, as shown. (a) Using the principle of
conservation of mechanical energy, find the speed of the 1.50 kg ball just before
impact.

So, KEi + PEi = KEf + PEf
1/2mv^2 + mgh [initial] = 1/2mv^2 + mgh [final]

The answer is 5.56m/s, but I'm getting something different. Can anyone walk me through the process? I'm not sure where I'm going wrong. Any help is greatly appreciated.

-
By the law of energy conservation:-
=>[KE+PE]initial = KE(final)
=>1/2mu^2 + mgh = 1/2mv^2
=>v^2 = u^2 + 2gh
=>v^2 = (5)^2 + 2 x 9.8 x 0.3
=>v = √30.88
=>v = 5.56 m/s

-
v^2 = u^2 + 2gh = 30.88 ---> v = 5.557
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