Question

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. If 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.

Answer 1

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

                   

Answer 2

The mass of the boat is 125 kg

The speed of the person is 1.75 m/s

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Further explanation

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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Given:

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

Asked:

mass of boat = m_b = ?

speed of the person = v_p = ?

Solution:

We will use Conservation of Momentum Law as follows:

First Case:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

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

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

$0 = -50v_p + 0.70m_b$

$50v_p = 0.70m_b$

$m_b = \frac{500}{7} v_p$ → Equation A

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Second Case:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

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

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

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

$100v_p = (50 + m_b)$

$100v_p = 50 + \frac{500}{7} v_p$ ← Equation A

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

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

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

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

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

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

$m_b = 125 \texttt{ kg}$

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Learn more

• 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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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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