Chapter 1: Q82P (page 269)

Two identical masses are released from rest in a smooth hemispherical bowl of radius R from the positions shown in Fig. P8.82. Ignore friction between the masses and the surface of the bowl. If the masses stick together when they collide, how high above the bottom of the bowl will they go after colliding?

Short Answer

Expert verified

The raised height for stuck blocks is $\frac{R}{4}$ .

Step by step solution

Determination of magnitude of the velocity of mass before the collisionGiven Data:

The height of the top mass from the bottom is h = R

The mass of each block is: m

Concept

The velocity of the top block is calculated first, then find the combined velocity of stacked mass and use energy conservation to height raised by both blocks.

The speed of the top is given as:

$u = \sqrt{2 g h}$

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

$u = \sqrt{2 g (R)} u = \sqrt{2 g R}$

Apply the momentum conservation to find the speed of stuck blocks.

$m u = (m + m) v v = \frac{u}{2} v = \frac{\sqrt{2 g R}}{2} V = \sqrt{\frac{g R}{2}}$

Determination of height raised by combined mass

Apply the energy conservation to find raised height for sticked blocks.

$\frac{1}{2} (m + m) v^{2} = (m + m) g H H = \frac{v^{2}}{2 g}$

Substitute all the values in the above equation.

$H = \frac{{(\sqrt{\frac{g R}{2}})}^{2}}{2 g} H = \frac{R}{4}$

Therefore, the raised height for sticked blocks is $\frac{R}{4}$ .

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Short Answer

Step by step solution

Determination of magnitude of the velocity of mass before the collisionGiven Data:

Concept

Determination of height raised by combined mass

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