Question

A stationary body explodes into two fragments of mass m1 and m2 the momentum of one fragment if be the minimum energy of explosion is

Answer 1

energy conservation

${p}^{1} + {p}^{2} = 0$
so , Energy
$= \frac{ {p}^{2} }{2m}$
so ,
$\frac{ {p}^{2} }{2 {m}^{1} } + \frac{ {p}^{2} }{ {2m}^{2} }$

so from this we get
$\frac{ {p}^{2}( {m}^{1} \: + {m}^{2} ) }{2 {m}^{1} {m}^{2} }$
note here
${m}^{2} = \: is \: not \: m \: square \: it \: is \: mass \: of \: object \: 2$
of object 2
pls. follow and mark it as brainlist

Answer 2

Answer:

Explanation:

Since the initial momentum of the system is zero. So, according to the conservation of linear momentum, final momentum of the system is zero. Thus momentum of the second fragment is also p.

Kinetic energy of first fragment  E1​=2m1​p2​

Kinetic energy of second fragment  E2​=2m2​p2​

Minimum energy of explosion   E=E1​+E2​=2m1​m2  =​p2/(m1​+m2​)