Question

A cart loaded with sand having total mass $900\text{kg}$ moves on a straight horizontal road under the action of a force $60\text{N}$. If cart is starting from rest and sand spills through a small hole at the bottom of cart at a rate of $0.25\text{kg/s},$ what will be the speed of cart after $10$ $\text{minutes}$?
Given: $g=10{\text{m/s}}^2,\ln \left(\frac{5}{6}\right)\approx -0.2$

• A. $50\text{m/s}$
• B. $68\text{m/s}$
• C. $24\text{m/s}$
• D. $48\text{m/s}$
  

Answer 1

The correct option is D $48\text{m/s}$


Initial mass of cart $m_i=900\text{kg}$
Rate of spilling of sand $=0.25\text{kg/s}$

$\Rightarrow$ Rate of change of mass of cart:
$\therefore \frac{dm}{dt}=0.25\text{kg/s}$

Total mass of sand spilled after time $t\text{s}$ is $0.25t$

Hence mass of cart after time $t\text{s}$ :
$m_f=(900-0.25t)\text{kg}$

From Newton's $2^{nd}$ law of motion:
$F=m\frac{dv}{dt}$
where $m=$ mass of cart at any given time
$\Rightarrow 60=(900-0.25t)\frac{dv}{dt}$
$\Rightarrow 60{\int }_{t=0}^{t=600}\frac{dt}{(900-0.25t)}={\int }_{v=0}^{v=v}dv$
where the speed of cart is $v$ after time $10\text{min}\text{or}600\text{s}$ ; at time $t=0,v=0$
$\Rightarrow v=60{\int }_0^{600}\frac{dt}{(900-0.25t)}=\frac{60}{-0.25}{\left[\ln \right(900-0.25t\left)\right]}_0^{600}$
$\therefore v=-240{\left[\ln \right(900-0.25t\left)\right]}_0^{600}$
$\therefore v=-240\times \left[\ln \right(900-150)-\ln (900-0\left)\right]$
$\Rightarrow v=-240\times \ln \left(\frac{750}{900}\right)=-240\times \ln \left(\frac{5}{6}\right)$
$\therefore v=48\text{m/s}$