A stationary body explodes into three fragments of masses $m_{1}$ , $m_{2}$ and $m_{3}$ . If the momentum of one fragment is p and one fragment of mass $m_{3}$ remains at rest, the energy of explosion is:

\frac{p^{2}}{2 (m_{1} + m_{2})}

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

\frac{p^{2}}{2 \sqrt{m_{1} m_{2}}}

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

\frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}

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

\frac{p^{2}}{2 (m_{1} - m_{2})}

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

Open in App

Solution

The correct option is D $\frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$
Given $, p$ is momentum and $m_{3}$ is at rest.
$p = m_{1} v_{1} = m_{2} v_{2}$
Energy of explosion= $\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{p^{2}}{2 m_{1}} + \frac{p^{2}}{2 m_{2}} = \frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$

Suggest Corrections

Similar questions

A stationary body explodes into three fragments of masses $m_{1}, m_{2}$ and $m_{3}$ . If momentum of one fragment is $p$ and one fragment of mass $m_{3}$ remains at rest, the energy of explosion is:

Q. A stationary body explodes into two fragments of masses

m_{1}

and

m_{2}

. If momentum of one fragment is p, the energy of explosion is :

Q. A stationary body explodes into two fragments of masses

m_{1}

and

m_{2}

. If momentum of one fragments p, the minimum energy of explosion is

Q. A stationary body explodes into two fragments of masses

m_{1}

and

m_{2}

. If momentum of one tangent is

p

, the minimum energy of explosion is

Q. A stationary bomb explodes into two fragments of masses

m_{1}

and

m_{2}

. IF the momentum of these fragments are

p_{1}

and

p_{2}

then what will be the ratio of their kinetic energies?

Join BYJU'S Learning Program

A stationary body explodes into three fragments of masses m1,m2 and m3. If the momentum of one fragment is p and one fragment of mass m3 remains at rest, the energy of explosion is:

The correct option is D p2(m1+m2)2m1m2Given,p is momentum and m3 is at rest.p=m1v1=m2v2Energy of explosion=12m1v21+12m2v22=p22m1+p22m2=p2(m1+m2)2m1m2

A stationary body explodes into three fragments of masses $m_{1}$ , $m_{2}$ and $m_{3}$ . If the momentum of one fragment is p and one fragment of mass $m_{3}$ remains at rest, the energy of explosion is: