A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. the energy released during the process is :

Question

A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. the energy released during the process is : (A) (3/5)mv2 (B) (5/3)mv2 (C) (3/2)mv2 (D) (4/3)mv2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/60ee4ce8168c7f0c60b285b705b4288e-.png" /> 3m vecv + mv hati + mv hatj = 0 ⇒ vecv = -(v/3) hati - (v/3) hatj | vecv| = √2 (v/3) Energy released = (1/2) mv2 + (1/2) mv2 + (1/2) (3m) ((2v2/9)) =(4/3) mv2

Q. A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. the energy released during the process is :

$\frac{3}{5} m v^{2}$

$\frac{5}{3} m v^{2}$

$\frac{3}{2} m v^{2}$

$\frac{4}{3} m v^{2}$

$3 m v + m v \hat{i} + m v \hat{j} = 0$
$\Rightarrow v = - \frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}$
$∣ v ∣ = 2 \frac{v}{3}$
Energy released
$= \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} (3 m) (\frac{2 v ^{2}}{9})$
$= \frac{4}{3} m v^{2}$

Q. A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. the energy released during the process is :

53​mv2

35​mv2

23​mv2

34​mv2

3mv+mvi^+mvj^​=0 ⇒v=−3v​i^−3v​j^​ ∣v∣=2​3v​ Energy released =21​mv2+21​mv2+21​(3m)(92v2​) =34​mv2

$\frac{3}{5} m v^{2}$

$\frac{5}{3} m v^{2}$

$\frac{3}{2} m v^{2}$

$\frac{4}{3} m v^{2}$

$3 m v + m v \hat{i} + m v \hat{j} = 0$
$\Rightarrow v = - \frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}$
$∣ v ∣ = 2 \frac{v}{3}$
Energy released
$= \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} (3 m) (\frac{2 v ^{2}}{9})$
$= \frac{4}{3} m v^{2}$