While lauching a rocket of mass `2xx 10^(4)kg` a force of ` 5 xx 10^(5) N` is applied for `20s` Calculate the velocity attained by the rocket at the end of `20s` .

Answer

`500m//s` .

Explanation

Here, `m = 2 xx 10^(4)kg, F = 5 xx 10^(5) N, t = 20 s`
`upsilon = ?, u = 0`
As Impulse = change in momentum
`:. F xx t = m (upsilon - u) = m upsilon`
`upsilon = (F xxt)/(m)= (5xx 10^(5) xx 20)/(2 xx10^(4)) = 500m//s` .