In a typical Indian Buggy (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. A buggy of mass $200 kg$ is moving at a speed of $10 km/h^{ }$. A person running at the speed of $4km/h$ jumps and sits on the wooden plate of the buggy., the person sits on the wooden plate. If the mass of the person is $25 Kg$, what will be the new velocity of the buggy in km/h?

 [[1]]

Using the formula for the law of conservation of linear motion in accordance with the mass of the buggy and the boy with respect to the buggy and boy's velocity, it is possible to determine the new velocity of the buggy in the situation stated above.
Formula used:
The rule of conservation of linear motion's formula yields the buggy's new speed;

$m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)\times V$

Where m₁ stands for the buggy's mass, m₂ for the boy's mass, v₁ for the bugghy's velocity, v₂ for the boy's velocity, and V for the buggy's ultimate new velocity.

The data given in the problem are;
The mass of the buggy is, $m_1=200Kg$.
The mass of the person is, $m_2=25Kg$.
The speed by which the buggy is moving is, $v_1=10Km{h}^{-1}$.
The speed by which the person is moving is, $v_2=4Km{h}^{-1}$.
Let the buggy and boy form a system.
The new velocity of the buggy is given by the formula of the law of conservation of linear motion;
So, the initial momentum of the system is equal to the final velocity.
$\Rightarrow m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)\times V$
Where, $V$ is the final new velocity of the buggy.
For,
$M_1=m_1{v}_1+m_2{v}_2$
Substituting the values for the mass of the buggy and the person and the values of the velocity of the buggy and the person in the above equation.
$\Rightarrow M_1=\left(200\times 10\right)+\left(25\times 4\right)$
On simplification,
$\Rightarrow M_1=2000+100$
$\Rightarrow M_1=2100$ (1)
For,
$M_2=\left(m_1+m_2\right)\times V$
Substituting the values for the mass of the buggy and the person and the values of the velocity of the buggy and the person in the above equation.
$\Rightarrow M_2=\left(200+25\right)\times V$
$\Rightarrow M_2=225V$ (2)
Since the final and the initial momentum acting on the body are equal.
$\Rightarrow M_1=M_2$
By equating the (1), (2) equations we get
$2100=225V$
Since we only need the final new velocity of the buggy;
$\Rightarrow V={\displaystyle \frac{2100}{225}}$
On simplification,
$\Rightarrow V={\displaystyle \frac{28}{3}}$
$\Rightarrow V=9.3 kmph$

∴ The final new velocity of the buggy is given as $V=9.3 kmph$.