The loss of kinetic energy during the above process is

\frac{I ω^{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{I ω^{2}}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{I ω^{2}}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{I ω^{2}}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{I ω^{2}}{3}$
Momentum conservation between the two situation-
$I (2 ω) + 2 I (ω) = I ω_{1} + 2 I ω_{1} 4 I ω = 3 I ω_{1} \Rightarrow ω_{1} = \frac{4 ω}{3}$
Initial K.E $= \frac{1}{2} I (2 ω^{2}) + \frac{1}{2} (2 I) ω^{2} = 3 I ω^{2}$
Final K.E $= \frac{1}{2} I (ω_{1})^{2} + \frac{1}{2} (2 I) ω_{1}^{2} = \frac{8 I ω^{2}}{3}$
Loss of K.E $= 3 I ω^{2} - \frac{8 I ω^{2}}{3} = \frac{I ω^{2}}{3}$

Suggest Corrections

Similar questions

If $ω$ is imaginary cube root of unity, then arg(i $ω$ ) + arg(i $ω^{2}$ ) =

If $ω$ = $\frac{- 1 + i \sqrt{3}}{2}$ , then $Z_{1}$ = $\frac{- \sqrt{3} - i}{2}$ and $Z_{2}$ = $\frac{\sqrt{3} - i}{2}$ can be expressed in terms of $ω$ and $ω^{2}$ as

Q. If

\sqrt{- 1} = i,

and

ω

is a nonreal cube root of unity then the value of

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & ω^{2} & 1 + i + ω^{2} - i & - 1 & - 1 - i + ω 1 - i & ω^{2} - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

is equal to

Q. The loss in kinetic energy of the system during collision is

Q. During "inelastic collision "
a) There is a loss of kinetic energy.
b) Some of the kinetic energy is used to deform the body.
c) Some of the kinetic energy is liberated as heat.
d) There is a loss of mass energy.

Join BYJU'S Learning Program

The loss of kinetic energy during the above process is

The correct option is B Iω23Momentum conservation between the two situation-I(2ω)+2I(ω)=Iω1+2Iω14Iω=3Iω1⇒ω1=4ω3Initial K.E=12I(2ω2)+12(2I)ω2=3Iω2Final K.E=12I(ω1)2+12(2I)ω21=8Iω23Loss of K.E=3Iω2−8Iω23=Iω23