Question

Two balls having masses m and 2m are fastened to two light strings of same length l which is held horizontally[say x axis, with the mass 2m at (l,0) and mass m at (-l,0)]. The strings are kept fixed at origin. This system is released from rest. Collision between the balls is elastic in nature.
a- find the velocities of the balls just after their collision and
b- how high will the balls rise after collision ?

Answer 1

Answer on Question #51469, Physics, Mechanics | Kinematics | Dynamics

Two balls having masses $m$ and $2m$ are fastened to two light strings of same length $l$ which is held horizontally [say $x$ axis, with the mass $2m$ at $(l,0)$ and mass $m$ at $(-l,0)$]. The strings are kept fixed at origin. This system is released from rest. Collision between the balls is elastic in nature.

a- find the velocities of the balls just after their collision and

b- how high will the balls rise after collision?

Solution:

As a ball begins moving, its potential energy is converted to kinetic energy.

$$m g l = \frac{m v^{2}}{2}$$

Thus, the velocity of first and second ball is the same and equal

$$v = \sqrt{2 g l}$$

The equation that denotes the conservation of momentum is:

$$m_{1} v_{1i} - m_{2} v_{2i} = - m_{1} v_{1f} + m_{2} v_{2f}$$

where, $m_{1} = m$ mass of object or ball 1

$m_{2} = 2m$ mass of ball 2

$v_{1i} = v$ initial velocity of ball 1

$v_{2i} = v$ initial velocity of ball 2

$v_{1f} =$ final velocity of ball 1

$v_{2f} =$ final velocity of ball 2

$$2 m v - m v = 2 m v_{2f} + m v_{1f}$$ $$v = v_{1f} + 2 v_{2f}$$

The kinetic energy conservation formula is

$$\frac{m v^{2}}{2} + \frac{2 m v^{2}}{2} = \frac{m v_{1f}^{2}}{2} + \frac{2 m v_{2f}^{2}}{2}$$ $$3 v^{2} = v_{1f}^{2} + 2 v_{2f}^{2}$$

Substituting

$$v_{2f} = \frac{v - v_{1f}}{2}$$ $$3 v^{2} = v_{1f}^{2} + 2 \left(\frac{v - v_{1f}}{2}\right)^{2}$$ $$3 v^{2} = v_{1f}^{2} + 2 \left(\frac{v^{2} - 2 v v_{1f} + v_{1f}^{2}}{4}\right)$$ $$3 v^{2} = v_{1f}^{2} + \frac{v^{2}}{2} - v_{1f} + \frac{v_{1f}^{2}}{2}$$ $$6 v^{2} = 2 v_{1f}^{2} + v^{2} - 2 v v_{1f} + v_{1f}^{2}$$ $$3 v_{1f}^{2} - 2 v v_{1f} - 5 v^{2} = 0$$ $$v_{1f} = \frac{2 v \pm \sqrt{4 v^{2} - 4 * 3 * (-5 v^{2})}}{6} = \frac{2 v \pm v \sqrt{64}}{6} = \frac{5 v}{3} \text{ or } -v$$$$v _ {2 f} = - \frac {1}{3} v \text{ or } v$$

We choose set of solution

$$v _ {1 f} = \frac {5 v}{3} = \frac {5}{3} \sqrt {2 g l}$$ $$v _ {2 f} = - \frac {1}{3} \sqrt {2 g l}$$

b.

$$h _ {1} = \frac {v _ {1 f} ^ {2}}{2 g} = \frac {2 5 * 2 g l}{9 * 2 g} = \frac {2 5}{9} l$$ $$h _ {2} = \frac {v _ {2 f} ^ {2}}{2 g} = \frac {2 g l}{9 * 2 g} = \frac {l}{9}$$

Answer: a. $v_{1f} = \frac{5}{3}\sqrt{2gl}; v_{2f} = -\frac{1}{3}\sqrt{2gl}$

b. $h_1 = \frac{25}{9} l; h_2 = \frac{l}{9}$

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