A 0.255 kg catcher is attached by a 45.0 cm long massless cable to a frictionles

Question

A 0.255 kg catcher is attached by a 45.0 cm long massless cable to a frictionless axle. The catcher is initially at rest. A 0.144 kg ball with a radius of 0.0875 m rolls without slipping from a height of 1.25 m and is captured by the catcher. For a solid ball, I = (2/5)MR^2. (a) What is the speed of the ball just before the ball is caught? (b) What is linear speed of the ball and catcher immediately after the ball is caught? Only linear momentum is conserved in this collision. (c) If the cable is shortened to 21.5 cm, what would the linear speed of the ball and launcher be?

Explanation / Answer

(a) Because the ball "rolls without slipping," it has rotational KE as well as translational KE when it reaches the catcher.

mgh = ½mv² + ½I²

We're told I = ½mR² and "without slipping" means that = v / R, so

mgh = ½mv² + ½(2/5)mR²(v/R)² = ½mv² + (1/5)mv² = 0.7mv²

mass m cancels, and

v = (gh / 0.7) = (9.8m/s² * 1.25m / 0.7) = 4.18 m/s

(b) Now conserve momentum:

0.144kg * 4.18m/s + 0 = (0.144 + 0.255)kg * V

v=1.51

(c)1.51

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