Question

A man weighing 80kg is standing at the centre of a flat boat and is 20m away from the shore.He walks 8m on the boat towards the shore and then halts.The boat weighs 200kg.How far is he from the shore at the end of this time?

A)11.2m

B)13.8m

C)14.3m

D)15.4m

Answer 1

Answer:

Explanation:

Correct answer is 14.3m

Answer 2

The is man 20 - 5.7 = 14.3 m away from the shore at the end.  

Explanation:

The center of mass of the system is given as

$X_cm=\frac{m_b_o_a_tX_1+m_m_a_nX_2}{m_b_o_a_t+m_m_a_n}$

As per question, boy and boat moving in opposite direction then the displacement of the center of mass is zero. Therefore above equation becomes

$0=m_b_o_a_tX_1+m_=m_m_a_nX_2$

Substitute the given values, we get

$200x+80(x-8)$

x = 2.3 m

Since the man walks 8 m on the boat towards the shore and then halts so displacement of the man = 8-2.3 = 5.7.

Thus, the man is 20 - 5.7 = 14.3 m away from the shore at the end.  

Option (c) is correct.

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