A ball of mass $5 k g$ is suspended from a chord of length $l = 0.5 m$ is attached to a cart $A$ of mass $2.5 k g$ , which slides freely on a frictionless horizontal track as shown in figure. The ball given an initial horizontal velocity $1 m / s$ relative to the $g r o u n d$ while the cart is at rest.
Find
(a) The velocity of the ball when string makes $m i n i m u m$ angle with vertical
(b) The maximum vertical distance through which the ball will rise.

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Solution

(a) At time of horizontal velocity of ball is same of cart velocity .
By linear momentum conservation
$5 x_{1} = 5 x v + 2.5 v \frac{5}{7.5} = v v = \frac{2}{3} m / s$
(b) By energy conservation theorem
$\frac{1}{2} 5 (1)^{2} = \frac{1}{2} 5 \times v^{2} + \frac{1}{2} 2.5 \times v^{2} + 5 g (h) 5 = \frac{20}{9} + \frac{10}{9} + 10 g h 5 - \frac{30}{9} = 10 \times 10 h \frac{15}{9} = 100 h h = \frac{5}{300} m = \frac{5}{3} c m$

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