The bob of a simple pendulum of mass and total energy textE textk will have maximum linear momentum equal to:

Question

The bob of a simple pendulum of mass and total energy textE textk will have maximum linear momentum equal to: (A)  textMEk2  (B)  text2mEk  (C)  √( text2Ek/m)  (D)  √ text2mEk

Tardigrade Admin · Accepted Answer

p max =mv max ?(i) Ek=(1/2)mvm2 ?(ii) Squaring equation (i), we get, p2=m2 upsilon max 2=m⋅ m upsilon max 2=2Ekm hence p=√2mEk

Q. The bob of a simple pendulum of mass and total energy $E_{k}$ will have maximum linear momentum equal to:

$ME_{k}^{2}$

$2mE_{k}$

$\frac{2E _{k}}{m}$

$2mE_{k}$

$p_{m a x} = m v_{m a x}$ ?(i) $E_{k} = \frac{1}{2} m v_{m}^{2}$ ?(ii) Squaring equation (i), we get, $p^{2} = m^{2} υ_{m a x}^{2} = m \cdot m υ_{m a x}^{2} = 2 E_{k} m$ hence $p = 2 m E_{k}$

Q. The bob of a simple pendulum of mass and total energy Ek​ will have maximum linear momentum equal to:

MEk2​

2mEk​

m2Ek​​​

2mEk​​

pmax​=mvmax​ ?(i) Ek​=21​mvm2​ ?(ii) Squaring equation (i), we get, p2=m2υmax2​=m⋅mυmax2​=2Ek​m hence p=2mEk​​

Q. The bob of a simple pendulum of mass and total energy $E_{k}$ will have maximum linear momentum equal to:

$ME_{k}^{2}$

$2mE_{k}$

$\frac{2E _{k}}{m}$

$2mE_{k}$

$p_{m a x} = m v_{m a x}$ ?(i) $E_{k} = \frac{1}{2} m v_{m}^{2}$ ?(ii) Squaring equation (i), we get, $p^{2} = m^{2} υ_{m a x}^{2} = m \cdot m υ_{m a x}^{2} = 2 E_{k} m$ hence $p = 2 m E_{k}$