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3 years ago

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After a 50-kg person steps from a boat onto the shore, the boat moves away with a speed of 0.70 m/s with respect to the shore. I

f the same person steps from the shore onto the boat, the person and the boat move away from the shore with the speed of 0.50 m/s. Assume that the initial speed of the person with respect to the shore is the same in both cases and that the person is moving in a horizontal direction. Determine the Mass of the boat and the speed of the person.

2 answers:

Svetradugi [14.3K]3 years ago

7 0

Answer:

M=125 kg

v=1.75 m/s

Explanation:

From the law of linear momentum

P =mv

Case 1 50*V =M* 0.7 equation 1

50*V =(M+50)* 0.5 equation 2

equating 1 and 2

M* 0.7=(m+50)* 0.5

0.2 M= 25

M=125 kg

Putting value of M in equation 1

50*V =125*0.7

V=1.75 m/s

KATRIN_1 [288]3 years ago

5 0

The mass of the boat is 125 kg

The speed of the person is 1.75 m/s

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<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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<u>Given:</u>

mass of person = m_p = 50 kg

final speed of the boat in first case = v_{b1} = 0.70 m/s

final speed of the boat in second case = v_{b2} = 0.50 m/s

<u>Asked:</u>

mass of boat = m_b = ?

speed of the person = v_p = ?

<u>Solution:</u>

<em>We will use </em><em>Conservation of Momentum Law </em><em>as follows:</em>

<u>First Case:</u>

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

$m_pu_p + m_bu_b = m_pv_p + m_bv_{b1}$

$50(0) + m_b(0) = -50v_p + m_b(0.70)$

$0 = -50v_p + 0.70m_b$

$50v_p = 0.70m_b$

$m_b = \frac{500}{7} v_p$ → <em>Equation A</em>

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<u>Second Case:</u>

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

$m_pv_p + m_bu_b = (m_p + m_b)v_{b2}$

$50v_p + m_b(0) = (50 + m_b)(0.50)$

$50v_p = (50 + m_b)(0.50)$

$100v_p = (50 + m_b)$

$100v_p = 50 + \frac{500}{7} v_p$ ← <em>Equation A</em>

$100v_p - \frac{500}{7} v_p = 50$

$\frac{200}{7}v_p = 50$

$v_p = \frac{7}{4} \texttt{ m/s}$

$v_p = 1.75 \texttt{ m/s}$

$\texttt{ }$

$m_b = \frac{500}{7} v_p$

$m_b = \frac{500}{7}(1.75)$

$m_b = 125 \texttt{ kg}$

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<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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