Question

A 500 $\text{kg}$ boat has an initial speed of $10\text{ m}{{\text{s}}^{-1}}$ as it passes under a bridge. At that a 50 $\text{kg}$ man jumps straight down into the boat from the bridge. The speed of the boat after the man and boat attains a common speed is:
(A) $\dfrac{100}{11}\text{ m}{{\text{s}}^{-1}}$
(B) $\dfrac{10}{11}\text{ m}{{\text{s}}^{-1}}$
(C) $\dfrac{50}{11}\text{ m}{{\text{s}}^{-1}}$
(D) $\dfrac{5}{11}\text{ m}{{\text{s}}^{-1}}$

Answer 1

One of the most powerful law & in physics is the law of momentum conservation. It is stated as for a collision occurring between object 1 and object 2, the total momentum of two objects before the collision is equal to the total momentum of the two objects after collision.
${{\text{P}}_{\text{i}}}={{\text{P}}_{\text{f}}}$

Complete step by step solution
In this mass of the boat ${{\text{m}}_{1}}=\text{500 kg}$
Speed of the boat u $=10\text{ m}{{\text{s}}^{-1}}$
Mass of man ${{\text{m}}_{2}}=50\text{ kg}$
We have to find out the speed of the boat after the man jumps into it.
According to the law of conservation of momentum, the initial momentum should be equal to final momentum. i.e.
${{\text{P}}_{\text{i}}}={{\text{P}}_{\text{f}}}$
We know that, the momentum is given by
P=m v, where m=mass of the body
v=velocity of the body
In initial condition
${{\text{P}}_{\text{i}}}$ is given by
${{\text{P}}_{\text{i}}}={{\text{m}}_{1}}\text{ u}$
${{\text{m}}_{1}}=$ mass of the boat
u=Velocity of the boat
Put the value, and initial momentum is given by:
For Boat
P $=500\times 10$
$=5000\text{ kg m/s}$
For man,
P $=50\times 0=0$
$\therefore {{\text{P}}_{\text{i}}}=5000\text{ kg m/s}$
Similarly in case of final condition, ${{\text{P}}_{\text{f}}}$ is given by:
${{\text{P}}_{\text{f}}}=\left( \text{M}+\text{m} \right)\text{ v}$ , where M=mass of man
m=mass of the boat
v=final velocity
$\begin{aligned} & {{\text{P}}_{\text{f}}}=\left( 500+50 \right)\text{ v} \\\ & \text{ }=550\text{ v} \\\ \end{aligned}$
According to law of conservation of moment
${{\text{P}}_{\text{i}}}={{\text{P}}_{\text{f}}}$
$\therefore \text{ 5000}=\text{550 v}$
$\begin{aligned} & \text{v}=\dfrac{5000}{550} \\\ & \text{ }=\dfrac{500}{55} \\\ & \text{ }=\dfrac{100}{11} \\\ \end{aligned}$
$\text{v}=\dfrac{100}{11}\text{ m/s}$
So speed of the boat after man jumps into it is $\dfrac{100}{11}\text{ m/s}$
Therefore the option (A) is correct.

Note
This is a very important law in collision. With the help of this law, we can find out the mass of bodies and their velocities. This law is very useful in our daily life. One of the applications of the conservation of momentum is the launching of rockets, airbags in automobiles, a device where, when one ball is lifted and then let go, the ball on the other end of a row of balls will push upward etc.