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A body of rest mass m_0 collides perfectly inelastically at a speed of 0.8c with anotherbody 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 . |
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Answer» Solution :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 therest 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 therest 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 therest 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 therest 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 therest 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 = 8/3m_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 restmasses of the individual bodies |
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