A body of rest mass `m_0` collides perfectly inelastically at a speed of 0.8c with another body of equal rest mass kept at rest . Calculate the common speed of the bodies after the collision and the rest mass of the combined body .

Explanation

The linear momentum of the first body
`= m_0 v / (sqrt 1 - v ^(2) / c^(2)) = m_0 xx 0.8c / 0.6 `
` = 4 / 3 m_0c.`This should be the total linear momentum after the collision. If the rest mass of the combined body is `M_0` and it moves at speed v'
`M_0 v' / (sqrt 1 - v'^(2) / c^(2)) = 4 / 3 m_0 c`
The energy before the collison is
`m_0 / (sqrt 1 -v '^(2) / c^(2)) c^(2) + m_0 c^(2) = m_0 C^(2) (1 / 0.6 +1)`The linear momentum of the first body
`= m_0 v / (sqrt 1 - v ^(2) / c^(2)) = m_0 xx 0.8c / 0.6 `
` = 4 / 3 m_0c.`This should be the total linear momentum after the collision. If the rest mass of the combined body is `M_0` and it moves at speed v'
`M_0 v' / (sqrt 1 - v'^(2) / c^(2)) = 4 / 3 m_0 c`
The energy before the collison is
`m_0 / (sqrt 1 -v '^(2) / c^(2)) c^(2) + m_0 c^(2) = m_0 c^(2) = m_0 c_^(2) (1 / 0.6 +1) ` The linear momentum of the first body
`= m_0 v / (sqrt 1 - v ^(2) / c^(2)) = m_0 xx 0.8c / 0.6 `
` = 4 / 3 m_0c.`This should be the total linear momentum after the collision. If the rest mass of the combined body is `M_0` and it moves at speed v'
`M_0 v' / (sqrt 1 - v'^(2) / c^(2)) = 4 / 3 m_0 c`
The energy before the collison is
`m_0 / (sqrt 1 -v '^(2) / c^(2)) c^(2) + m_0 c^(2) = m_0 C^(2) (1 / 0.6 +1)` The linear momentum of the first body
`= m_0 v / (sqrt 1 - v ^(2) / c^(2)) = m_0 xx 0.8c / 0.6 `
` = 4 / 3 m_0c.`This should be the total linear momentum after the collision. If the rest mass of the combined body is `M_0` and it moves at speed v'
`M_0 v' / (sqrt 1 - v'^(2) / c^(2)) = 4 / 3 m_0 c`
The energy before the collison is
`m_0 / (sqrt 1 -v '^(2) / c^(2)) c^(2) + m_0 c^(2) = m_0 C^(2) (1 / 0.6 +1)`
`= 8/ 3 m_0c^(2). The linear momentum of the first body
`= m_0 v / (sqrt 1 - v ^(2) / c^(2)) = m_0 xx 0.8c / 0.6 `
` = 4 / 3 m_0c.`This should be the total linear momentum after the collision. If the rest mass of the combined body is `M_0` and it moves at speed v'
`M_0 v' / (sqrt 1 - v'^(2) / c^(2)) = 4 / 3 m_0 c`
The energy before the collison is
`m_0 / (sqrt 1 -v '^(2) / c^(2)) c^(2) + m_0 c^(2) = m_0 C^(2) (1 / 0.6 +1)`
`= 8/ 3 m_0 c^(2)`
The energy after collision is `M_0 c^(2) / (sqrt 1 - v'^(2) / c^(2) )`
Thus , `M_0 c^(2) / (sqrt 1 - v'^(2) / c^(2)) = 8 /3 m_0c^(2)`
Dividing (i) by (ii) ,
`v' / c^(2) = 1 / 2c or, v' = c / 2 .`
Putting this value of v' in (ii)
`M_0 = 8/3 m_0 (sqrt 1 - 1/ 4)`
or, `M_0 = 2.309 m_0`
The rest mass of the combined body is greater than the sum of the rest masses of the individual bodies