Question

A trolley of mass $100kg$, starting from rest, describes $100m$ in $10$ seconds. At that instant, i.e. at the commencement of the ${11}^{th}$ second, two packets, each of mass $12.5kg$, are gently placed on the trolley. How far (in $m$) does it move in the next $10$ seconds assuming that the forces on the trolley remain the same throughout?

Answer 1

Since $s=ut+\frac{1}{2}a{t}^2$
$100=0+\frac{1}{2}a100\therefore a=2m/se{c}^2$
The velocity at the end of 10 seconds = $v=u+at=0+2\times 20=20m/s$
$F=ma=100\times 2=200N$

When packets are dropped net mass of the system becomes $125\text{kg}$
During the next 10 seconds, acceleration = $\frac{200}{125}=1.6m/se{c}^2$
The velocity at the commencement of the ${11}^{th}$ second, $v^{\prime }$ is given by
$mu=m^{\prime }{v}^{\prime }$ (law of conservation of momentum)
$100\times 20=125\times v^{\prime }$
$\therefore v^{\prime }=\frac{100\times 20}{125}=16m/sec$
$\therefore$ The distance travelled in the next 10 seconds = $s^{\prime }=ut+\frac{1}{2}a{t}^2$
$=16\times 10+\frac{1}{2}\times 1.6\times 100=240m$