A truck of mass `2 xx 10^(4)kg` travelling at `1.5m//s` collindes with another truck of mass `3 xx 10^(4)kg` moving with a velocity of `2.5m//s` in opposite direction If the trucks couple on collision what is their velocity?

Answer

`-0.9m//s` .

Explanation

Here, `m_(1) = 2xx10^(4) kg, upsilon_(1) = 1.5 m//s`
`m_(2) = 3 xx 10^(4) kg, upsilon_(2) = - 2.5 m//s, upsilon = ?`
According to principle of conservation of linear momentum,
`(m_(1) + m_(2)) upsilon = m_(1) upsilon_(1) + m_(2) upsilon_(2)`
`(2 xx 10^(4)) upsilon = 2 xx 10^(4) xx 1.5 + 3 xx 10^(4) (-2.5)`
` 5 xx 10^(4) upsilon = 3xx 10^(4) upsilon = 3 xx 10^(4) - 7.5 10^(4) = - 4.5 xx 10^(4)`
`upsilon = - (4.5)/(5) = - 0.9 m//s`
Minus sign indicates velocity is along `upsilon_(2)` .