Question

A tank cart is filled with water as shown in the figure. The initial mass of the cart with water is $m_0=100\text{kg}$. At time $t=0$, a hole is made in the left wall of the cart and water starts spilling out from the cart, with a constant velocity of $v_r=10\text{m/s}$ with respect to the cart. The rate of ejection of water is $r=2\text{kg/s}$. Find the acceleration of the cart after time $t=5\text{s}$.

• A. $0.11{\text{m/s}}^2$
• B. $0{\text{m/s}}^2$
• C. $0.22{\text{m/s}}^2$
• D. $0.4{\text{m/s}}^2$

Answer 1

The correct option is C $0.22{\text{m/s}}^2$

Force of thrust on the cart
$F_t=-v_r\frac{dm}{dt}=-10\times 2=-20\text{N}$
(-ve sign indicates force is rightward)
Since there is no external force in horizontal direction,
$F_t=ma$

$\Rightarrow$ Acceleration of the cart $a=\frac{F_t}{m}$
$=\frac{F_t}{m_0-\mu t}=\frac{v_r\left(\frac{dm}{dt}\right)}{m_0-\mu t}$

At $t=5\text{s}$,
$a=\frac{10\times 2}{100-2\times 5}=0.22{\text{m/s}}^2$
$\therefore a=0.22{\text{m/s}}^2$