Question

Father (80 kg) and son (30 kg) are sitting at one
of the ends of a 4 m long boat (40 kg) standing
still on water. They start to shift slowly. Father
stopped at centre of boat but son stopped at other
end. Neglecting friction with water, how far does the
boat move on the water during the process?
(1) 1.87 m
(2) 1.5 m
(3) 1.25 m
(4) 1.2 m​

Answer 1

Answer:

The  distance traveled by the boat on the water is 1.5 m.

(2). is correct option.

Explanation:

Given that,

Mass of father = 80 kg

Mass of son = 30 kg

Mass of  boat = 40 kg

Length of boat = 4 m

Let A be the origin of entire motion.

We need to calculate the  distance traveled by the boat

Using formula of center of mass

$C_{m}=\dfrac{M_{1}\times d_{1}+M_{2}\times d_{2}+M_{3}\times d_{3}}{M_{1}+M_{2}+M_{3}}$

Where, $M_{1}$= Mass of father

$M_{2}$= Mass of boat

$M_{3}$= Mass of son

$d_{1}$= distance of father from the origin

$d_{2}$= distance from the origin

$d_{3}$= distance of son from the origin

Put the value into the formula

$C_{m}=\dfrac{60\times 0+40\times 2+30\times4}{60+40+30}$

$C_{m}=1.5\ m$

Hence, The  distance traveled by the boat on the water is 1.5 m.