A particle of mass 5m at rest suddenly breaks | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the velocity of the third fragment of mass $3m$ be $\vec{v}'$. According to the law of conservation of linear momentum,the initial momentum is

Vedclass · Accepted Answer

$\frac{4}{3} mv^{2}$

Vedclass · Answer

(D) Let the velocity of the third fragment of mass $3m$ be $\vec{v}'$. According to the law of conservation of linear momentum,the initial momentum is zero,so the final momentum must also be zero.$3m \vec{v}' + m v \hat{i} + m v \hat{j} = 0$$3m \vec{v}' = -mv \hat{i} - mv \hat{j}$$\vec{v}' = -\frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}$The magnitude of the velocity of the third fragment is $|\vec{v}'| = \sqrt{(-\frac{v}{3})^2 + (-\frac{v}{3})^2} = \sqrt{\frac{2v^2}{9}} = \frac{\sqrt{2}}{3} v$.The energy released is equal to the total kinetic energy of the fragments:$K = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (3m) |\vec{v}'|^2$$K = m v^2 + \frac{3}{2} m (\frac{2v^2}{9}) = m v^2 + \frac{1}{3} m v^2 = \frac{4}{3} m v^2$.

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

Similar Questions

An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally. Instantaneously after the firing,the ratio of kinetic energy of the artillery and that of the shell is:

If a shell fired from a cannon explodes in mid-air,then:

$A$ body of mass $M$ moving with velocity $V$ explodes into two equal parts. If one part comes to rest and the other part moves with velocity $v_{0}$,what would be the value of $v_{0}$?

$A$ body,whose momentum is constant,must have constant

$A$ bomb of mass $12\,kg$ at rest explodes into two fragments of masses in the ratio $1 : 3$. The $K.E.$ of the smaller fragment is $216\,J$. The momentum of the heavier fragment is (in $kg\cdot m/s$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module