Freezatron 9000

Level 2

Water of mass $m_{0}$ is completely filled in a hollow sphere of mass $m$ and radius $r$ . A small massless device, which can completely freeze the water in almost no time, is also put in it. The sphere is then released on the top of a rough inclined plane of height $h$ and angle of inclination $\theta$ .

When the ball is at half the original height, it has a speed of $v_{0}$ . The ball is instantly stopped there, the device is turned on, and let go again. The ball arrives at the bottom with a speed $v$ .

What is the value of $v_{0} - v$ ?

Details :

$h = 31.5 m , \theta = 30^{o} , m = 2.4 kg , m_{0} = 3 kg, r = 10cm$
There is enough friction to ensure rolling.

The answer is 1.185.

1 solution

Parth Sankhe
Oct 10, 2018

For the first half, the total kinetic energy = the loss in potential energy.

$\frac {1}{2}(m + m_{0})v_{0}^{2} + \frac {1}{2}Iw^{2} = mg\frac {h}{2}$

Here $I = \frac {2}{3}mr^{2}$ and $v_{0}=rw$

At the half mark, the water will freeze. As it will try to expand, it will press against the inner wall of the sphere and get latched onto it. Thus the water would turn into a solid sphere in a hollow sphere. Applying energy conservation again,

$\frac {5}{6}mv^{2} + \frac {7}{10}m_{0}v^{2} = mg\frac {h}{2}$

(The above 2 kinetic energies are the respective total kinetic energies in pure rolling of a hollow sphere and a solid sphere.)

Putting in the values would give $v = 14.402$ and $v_{0}=15.588$

Freezatron 9000

The answer is 1.185.

1 solution

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