Two pendulums each of length l are initially situated as shown in figure. The first pendulum is released and strikes the second. Assume that the collision is completely inelastic and neglect the mass of the string and any frictional effects. How high does the centre of mass rise after the collision?

Question

Two pendulums each of length l are initially situated as shown in figure. The first pendulum is released and strikes the second. Assume that the collision is completely inelastic and neglect the mass of the string and any frictional effects. How high does the centre of mass rise after the collision? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ba5c0c269262c4e614cf32f43e9a0d2c-.png" /> (A) d[(m1/(m1+m2))]2  (B) d=[(m1/(m1+m2))] (C) (d(m1+m2)2/m2) (D) d[(m2/(m1+m2))]2

Tardigrade Admin · Accepted Answer

Applying the law of conservation of momentum

m_{1} v_{1} = (m_{1} + m_{2}) V

where

v_{1} = 2 g d

is the velocity with which

m_{1}

collides with

m_{2}

We get:

V = \frac{m _{1}}{( m _{1} + m _{2} )} 2 g d

Now, let the centre of mass rise through a height

h

after collision.
In this case, the kinetic energy of

(m_{1} + m_{2})

system is converted into potential energy at maximum height

h .

\frac{1}{2} (m_{1} + m_{2}) V^{2} = (m_{1} + m_{2}) g h

\frac{1}{2} (m_{1} + m_{2}) {\frac{m _{1}}{m _{1} + m _{2}}}^{2} \cdot 2 g d

= (m_{1} + m_{2}) g h

h = d {\frac{m _{1}}{m _{1} + m _{2}}}^{2}

$d [\frac{m _{1}}{( m _{1} + m _{2} )}]^{2}$

$d = [\frac{m _{1}}{( m _{1} + m _{2} )}]$

$\frac{d ( m _{1} + m _{2} ) ^{2}}{m _{2}}$

$d [\frac{m _{2}}{( m _{1} + m _{2} )}]^{2}$

d[(m1​+m2​)m1​​]2

d=[(m1​+m2​)m1​​]

m2​d(m1​+m2​)2​

d[(m1​+m2​)m2​​]2

$d [\frac{m _{1}}{( m _{1} + m _{2} )}]^{2}$

$d = [\frac{m _{1}}{( m _{1} + m _{2} )}]$

$\frac{d ( m _{1} + m _{2} ) ^{2}}{m _{2}}$

$d [\frac{m _{2}}{( m _{1} + m _{2} )}]^{2}$