Question

In a typical Indian Bugghi (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 $10\,\,Km{h^{ - 1}}$. As it overtakes a schoolboy walking at a speed of $4\,\,Km{h^{ - 1}}$, the boy sits on the wooden plate. If the mass of the boy is \[25\,\, Kg\], what will be the new velocity of the buggy?

Answer 1

Hint: In the given above problem the new velocity of the buggy can be calculated by using the formula of the law of conservation of linear motion in accordance with the mass of the buggy and the boy with the relation to the velocity of the buggy and the boy.

Formula used:
The new velocity of the bugghi is given by the formula of law of conservation of linear motion;
${m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right) \times V$
Where, ${m_1}$ denotes the mass of the bugghi, ${m_2}$ denotes the mass of the boy, ${v_1}$ denotes the velocity of the bugghi, ${v_2}$ denotes the velocity of the boy, $V$ is the final new velocity of the bugghi.

Complete step by step answer:
The data given in the problem are;
The mass of the bugghi is, ${m_1} = 200\,\,Kg$.
The mass of the boy is, ${m_2} = 25\,\,Kg$.
The speed by which the bugghi is moving is, ${v_1} = 10\,\,Km{h^{ - 1}}$.
The speed by which the boy is moving is, ${v_2} = 4\,\,Km{h^{ - 1}}$.
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,
${F_1} = {m_1}{v_1} + {m_2}{v_2}$
Substituting the values for the mass of the buggy and the boy and the values of the velocity of the buggy and the boy in the above equation.
$\Rightarrow {F_1} = \left( {200 \times 10} \right) + \left( {25 \times 4} \right) $
On simplification,
$\Rightarrow {F_1} = 2000 + 100 $
$\Rightarrow {F_1} = 2100 $.................................(1)
For,
${F_2} = \left( {{m_1} + {m_2}} \right) \times V$
Substituting the values for the mass of the buggy and the boy and the values of the velocity of the buggy and the boy in the above equation.
$\Rightarrow {F_2} = \left( {200 + 25} \right) \times V $
$\Rightarrow {F_2} = 225\,\,V $................................(2)
Since the final and the initial forces acting on the body are equal.
$\Rightarrow {F_1} = {F_2}$
By equating the (1), (2) equation we get
$2100 = 225\,\,V$
Since we only need the final new velocity of the buggy;
$\Rightarrow V = \dfrac{{2100}}{{225}} $
On simplification,
$\Rightarrow V = \dfrac{{28}}{3} $
$\Rightarrow V = 9.3\,Kmph $

$\therefore$ The final new velocity of the bugghi is given as $V=9.3Kmph$.

Note:
The law of conservation of linear momentum states that the total momentum of the system is always preserved if no exterior force applies to an object or system of objects. That’s why we took the logic that the initial and final momentum are equal.