Question

67.) A boat of mass 40 kg is at rest. A dog of mass 4 kg
moves in the boat with a velocity of 10 m/s. What
is the velocity of boat?
A. 4 m/s
B. 2 m/s
C 8 m/s
D. 1 m/s​

Answer 1

Proper Question: A boat of mass 40 kg is at rest. A dog of mass 4 kg moves in the boat with a velocity of 10 m/s. What is the velocity of boat?

Provided that:

• Mass of boat = 40 kg
• Initial velocity of boat = 0 m/s
• Mass of dog = 4 kg
• Initial velocity of dog = 10 m/s
• Final velocity of dog = 0 m/s

To calculate:

• The final velocity of boat

Solution:

• The final velocity of boat = 1 m/s

Using concept:

${\small{\underline{\boxed{\sf{\rightarrow m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$] denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.

Required solution:

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 40(0) + 4(10) = 40(v_A) + 4(0) \\ \\ :\implies \sf 0 + 40 = 40(v_A) + 0 \\ \\ :\implies \sf 40 = 40(v_A) + 0 \\ \\ :\implies \sf 40 = 40(v_A) \\ \\ :\implies \sf 40/40 \: = v_A \\ \\ :\implies \sf 1 \: = v_A \\ \\ :\implies \sf v_A \: = 1 \: ms^{-1} \\ \\ :\implies \sf Velocity \: = 1 \: ms^{-1} \\ \\ :\implies \sf Velocity \: of \: boat \: = 1 \: ms^{-1}$