Question

a pendulum Bob of mass 10 ^ - 2 kg is raised to a height of 5 × 10 ^ - 2 metre and then released at the bottom of its swing it picks up a mass of 10 ^-3 kg .to what height will the combined mass rise. take g = 10 m/s^2​

Answer 1

Answer:

The combined mass will rise to the height of 0.041 m

Step-by-step explanation:

Mass of the bob of pendulum $M=10^{-2}$ Kg

Height upto which the bob is raised $h=5\times 10^{-2}$ m

Mass of the particle $m=10^{-3}$ Kg

When the bob will be at the bottom, its Potential Energy will have converted into Kinetic Energy

Therefore,

$Mgh=\frac{1}{2}Mv^2$

$\implies v=\sqrt{2gh}$

$\implies v=\sqrt{2\times 10\times 5\times 10^{-2}}$

$\implies v=\sqrt{1}=1$ m/s

When the bob picks the particle, there will be change in momentum

Let the velocity of the combined bob+particle be v'

Then

$Mv=(M+m)v'$

$\implies 10^{-2}\times 1=(10^{-2}+10^{-3})v'$

$\implies 10^{-2}=10^{-2}(1+0.1)v'$

$\implies v'=\frac{1}{1.1}$m/s

Let the height at which combined mass will rise is h' then

$\frac{1}{2}(M+m)v'^2=(M+m)gh'$

$\implies h'=\frac{1}{2g}\times v'^2$

$\implies h'=\frac{1}{2\times 10}\times (\frac{1}{1.1})^2$

$\implies h'=\frac{1}{20}\times\frac{100}{121}$

$\implies h'=\frac{5}{121}$

$\implies h'=0.0413$ m

Therefore, the combined mass will rise to the height of 0.041 m

Hope this helps.