Question #59aa4

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Question #59aa4

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Answer 1 · 2016-12-16T20:49:22

$\sqrt{2} v$ $sqrt2v^$

Explanation:

At height $h$ $h$ total energy of the cart $= K E + P E = \frac{1}{2} m v^{2} + m g h$ $=KE+PE=1/2mv^2+mgh$
where $m$ $m$ is mass of cart and $g$ $g$ is acceleration due to gravity.

After it looses its ( $50 %$ $50%$ ) half mass it climbs to height $2 h$ $2h$
Let velocity at that height $= v_{n}$ $=v_n$

At height $2 h$ $2h$ total energy of the cart $= \frac{1}{2} (\frac{m}{2}) v_{n}^{2} + (\frac{m}{2}) g (2 h)$ $=1/2(m/2)v_n^2+(m/2)g(2h)$
$= (\frac{m}{4}) v_{n}^{2} + m g h$ $=(m/4)v_n^2+mgh$

Assuming no loss of energy due to friction, and applying law of conservation of energy we get

$\frac{1}{2} m v^{2} + m g h = (\frac{m}{4}) v_{n}^{2} + m g h$ $1/2mv^2+mgh=(m/4)v_n^2+mgh$
$\Rightarrow \frac{v_{n}^{2}}{4} = \frac{v^{2}}{2}$ $=>v_n^2/4=v^2/2$
$\Rightarrow v_{n} = \sqrt{2} v$ $=>v_n=sqrt2v^$

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Question #59aa4

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