A cart loaded with sand having total mass m 0 = 1800 kg moves on a straight horizontal road starting from…

In this case, sand falls on the ground vertically and comes to rest. Hence, horizontal velocity of ejected sand with respect to ground is zero; so

$F{}_{\text{ext}}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathit{\text{mv}}\right)\text{or}F{}_{\text{ext}}\text{d}\mathit{\text{t}}=\mathrm{d}\left(\mathit{\text{mv}}\right)$

Which on integration gives

$F{}_{\text{ext}}{\int }_{t_1}^{t_2}\text{d}\mathit{\text{t}}=\int d\left(\mathit{\text{mv}}\right)$

$F{}_{\text{ext}}\left(t_2-t_1\right)=\left(m{}_2{v}_2-m{}_1{v}_1\right)\dots \left(\text{i}\right)$

The rate of leakage of sand, $K = \left(\raisebox{1ex}{$–\mathrm{d}m$}\!\left/ \!\raisebox{-1ex}{$\mathrm{d}t$}\right.\right)$ is constant which on integration gives

${\int }_{m{}_1}^{m{}_2}\mathrm{d}m=-K{\int }_{t_1}^{t_2}\mathrm{d}t$

$(m_2 – m_1) = –K(t_2 – t_1)$                     ...(ii)

Here,   $t_1= 0,, t_2 = t, v_1 = 0, v_2 = v$

From Eqns. (i) and (ii), 

$F_{\mathrm{ext}} .t = (m_0 – Kt)v$

$v=\frac{F{}_{\text{ext}}·t}{\left(m_0-Kt\right)}$

Substituting for$F_{\mathrm{ext}}, t, K$ and$m_0,$ we get

$v=\frac{120\times 20\times 60}{\left(1800-0\text{.}5\times 20\times 60\right)}=120 {\text{m s}}^{-1}$