Question

a boy of mass 25 kg jumps out of Rowing Boat of mass 300 kg on to the bank with the horizontal velocity of 3 metre per second with what velocity does the boat begin to move backward?

a) 0.214m/s
b) 0.124m/s
c) 4.6m/s
d) 2.64 m/s.

If anyone will solve it, I will make brainliest ​​

Answer 1

Explanation:

Solution :

Consider the board boy as a system. The exterN/Al forces on this system are a. weight of the system and b. normal contact force by the ice surface. Both these force are vertical and there is no exterN/Al force in horizontal directio. The horizontal component of linear momentum of the board+boy system is therefore constant. <br> If the board redcoils at a speed v <br>

<br> or,

<br> The boy and the board are separating wilth a rate <br>

Answer 2

Explanation:

$\sf \red {\underline{\underline{Given:-}}}$

• Mass of boy = 25kg

• Mass of boat = 300kg

• Horizontal velocity = 3m/s

$\sf \blue {\underline{\underline{To\:Find:-}}}$

• The velocity with which the boat begin to move backward.

$\sf \green{\underline{\underline{Solution:-}}}$

$\sf According \: \: to \: \: the \: \: law \: of \: \: conservation \: \: of \: momentum$

$\boxed{\sf \purple{ \implies m_{1} v_{1} + m_{2} v_{2} = 0 }}$

Here,

• $\sf m_{1} = mass \: of \: the \: boy$

• $\sf m_{1} = mass \: of \: the \: boat$

• $\sf v_{1} = horizontal \:velocity$

• $\sf v_{2} = velocity \: in \: backward\: direction$

$\sf { \underline{Substituting \: \: the \: \: values:-}}$

$\sf \implies m_{1} v_{1} + m_{2} v_{2} = 0$

$\sf \implies (25 \times 3) + 300 (v_{2} )= 0$

$\sf \implies 75 + 300 (v_{2} )= 0$

$\sf \implies 300 (v_{2} )= - 75$

$\sf \implies v_{2} = \dfrac{ - 75}{ \: \:300}$

$\sf \implies v_{2} = - 2.5$.....approx