# Question 59aa4

Dec 16, 2016

sqrt2v^

#### Explanation:

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

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

At height $2 h$ total energy of the cart$= \frac{1}{2} \left(\frac{m}{2}\right) {v}_{n}^{2} + \left(\frac{m}{2}\right) g \left(2 h\right)$
$= \left(\frac{m}{4}\right) {v}_{n}^{2} + m g h$

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

$\frac{1}{2} m {v}^{2} + m g h = \left(\frac{m}{4}\right) {v}_{n}^{2} + m g h$
$\implies {v}_{n}^{2} / 4 = {v}^{2} / 2$
=>v_n=sqrt2v^#