Question

A boy of mass $60Kg$ is standing over a platform of mass $40Kg$ placed over a smooth horizontal surface. He throws a stone of mass $1Kg$ with velocity $v=10m/s$ at an angle of ${45}^o$ with respect to the ground. Find the displacement of the platform (with boy) on the horizontal surface when the stone lands on the ground. ($g=10m/s$)

Answer 1

Given that,

Mass of boy $m_b=60kg$

Platform of mass $m_p=40kg$

Stone of mass $m_s=1kg$

Velocity of stone $u=10m/s$

Angle $\theta ={45}^0$

We know that,

Range of stone is

$x_{stone}=\frac{u^2\sin 2\theta }{g}$

$x_{stone}=\frac{{\left(10\right)}^2\sin {90}^0}{10}$

$x_{stone}=10m$

Now, the center of mass of the system of stone and (boy +platform) remains stationary

$x_{cm}=\frac{m_s\times x_s+(m_b+m_p)x_p}{m_s+m_b}=0$

$0=\frac{1\times 10+100x_p}{1+60}$

$10+100x_p=0$

$x_p=-0.1m$

$x_p=-10cm$

Negative sign indicates that displacement of platform along with boy is opposite to that of stone.

Hence, the displacement of the platform is $-10cm$