Question

17. A particle of mass m moving with velocity v
a simple pendulum of mass m and sticks to
maximum height attained by the pendulum will be?

Answer 1

Answer:

Given:

Mass of particle = m

Mass of Bob of Pendulum = m

Velocity of particle = v

To find:

How high the pendulum arises .

Concept:

Momentum will be conserved during the collision. Using this , we can find out the velocity of combined mass.

Then we will apply Conservation of Energy.

Calculation:

Conservation of Momentum :

$mv + 0 = (m + m)(v2)$

$= > mv = 2m(v2)$

$= > v2 = \dfrac{v}{2}$

Conservation the Energy :

$\frac{1}{2} (2m) {(v2)}^{2} = (2m)gh$

$= > \dfrac{1}{2} {( \dfrac{v}{2}) }^{2} = gh$

$= > h = \dfrac{ {v}^{2} }{8g}$

So final answer :

$\boxed{ \red{ \huge{h = \dfrac{ {v}^{2} }{8g} }}}$

Answer 2

Given

particle mass= m

intial velocity= $V_0$

To find

Height attained by pendulum

solution

m$V_0$=2mV

=>$V_0$=2V

=>$\frac{V_0}{2}$=V

$\boxed{\boxed{Kinetic\:energy=1/2mv^2}}$

Putting the value of v in kinetic energy

=$\frac{1}{2}$mv²

=$\frac{1}{2}$2m($\frac{V_0}{2}$)²

=$\frac{1}{2}$2m($\frac{V_0}{2}$

=$\frac{mv_0^2}{4}$

Equating it with potential energy

To getting height

$\boxed{\boxed{potential\:energy=mgh}}$

=>$\frac{mv_0^2}{4}$=mgh

=>$\frac{v_0^2}{4g}$=h