Two pendulum each of length `l` are initial 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 center of mass rise after the collision?

Options

A. `d [(m_(1))/((m_(1) + m_(2)))]^(2)`
B. `d [(m_(1))/((m_(1) + m_(2)))]`
C. `(d(m_(1) + m_(2))^(2))/(m_(2))`
D. `d [(m_(2))/((m_(1) + m_(2)))]^(2)`

Answer

A

Explanation

Applying the law of conservationof momentum
`m_(1) v_(1) = (m_(1) + (m_(2))V`
`v_(1) = sqrt(5gl)` is the velocity with which `m_(1)` collides with `m_(2)`. We get `V = (m_(1))/((m_(1) + m_(2))) sqrt(2gd)`
Now , let the center 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`
`(1)/(2) (m_(1) + m_(2)) V^(2) = (m_(1) + m_(2)) gh`
`(1)/(2) (m_(1) + m_(2)) {(m_(1))/(m_(1) + m_(2))} ^(2) . 2 gd = (m_(1) + m_(2)) gh`
`h = d {(m_(1))/(m_(1) + m_(2))} ^(2)`