Two people, one of mass 85 kg and the other of mass 55 kg, sit in a rowboat of mass 58 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.0 m apart from each other, now exchange seats. How far and in what direction will the boat move?

Short Answer

Expert verified

The distance moved by the boat towards the initial position of the 85-kg person is \(0.45\;{\rm{m}}\).

Step by step solution

01

Define center of mass (CM)

For a symmetrical object of uniform density, the CM is its geometric center. The velocity of the CM of a system of objects is the entire momentum divided by the system's total mass.

02

Given information

The mass of the first person is\({m_1} = 85\;{\rm{kg}}\).

The mass of the second person is\({m_2} = 55\;{\rm{kg}}\).

The mass of the boat is \(M = 58\;{\rm{kg}}\).

03

Calculate the center of mass of the system for the initial condition

The following figure shows the initial position of the masses.

Here,\({x_1} = 0\)is the x-coordinate of mass\({m_1}\),\({x_2} = 1.5\;{\rm{m}}\)is the x-coordinate of mass\(M\), and\({x_3} = 3.0\;{\rm{m}}\)is the x-coordinate of mass\({m_3}\).

The center of mass of the system can be calculated as shown below:

\(\begin{array}{l}{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{initial}}}} = \frac{{{m_1}{x_1} + M{x_2} + {m_2}{x_3}}}{{{m_1} + M + {m_3}}}\\{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{initial}}}} = \frac{{\left( {85\;{\rm{kg}}} \right)\left( 0 \right) + \left( {58\;{\rm{kg}}} \right)\left( {1.5\;{\rm{m}}} \right) + \left( {55\;{\rm{kg}}} \right)\left( {3.0\;{\rm{m}}} \right)}}{{\left( {85\;{\rm{kg}}} \right) + \left( {58\;{\rm{kg}}} \right) + \left( {55\;{\rm{kg}}} \right)}}\\{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{initial}}}} = 1.27\;{\rm{m}}\end{array}\)

04

Calculate the center of mass of the system for the final condition

When the seats of two people are exchanged, the following figure can be drawn.

Here,\({x_1} = 0\)is the x-coordinate of mass\({m_2}\),\({x_2} = 1.5\;{\rm{m}}\)is the x-coordinate of mass\(M\), and\({x_3} = 3.0\;{\rm{m}}\)is the x-coordinate of mass\({m_1}\).

The center of mass of the system can be calculated as shown below:

\(\begin{array}{c}{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{final}}}} = \frac{{{m_2}{x_1} + M{x_2} + {m_1}{x_3}}}{{{m_1} + M + {m_3}}}\\{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{final}}}} = \frac{{\left( {55\;{\rm{kg}}} \right)\left( 0 \right) + \left( {58\;{\rm{kg}}} \right)\left( {1.5\;{\rm{m}}} \right) + \left( {85\;{\rm{kg}}} \right)\left( {3.0\;{\rm{m}}} \right)}}{{\left( {85\;{\rm{kg}}} \right) + \left( {58\;{\rm{kg}}} \right) + \left( {55\;{\rm{kg}}} \right)}}\\{\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{final}}}} = 1.727\;{\rm{m}}\end{array}\)

05

Calculate the distance moved by the boat

\(\begin{array}{c}d = {\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{final}}}} - {\left( {{x_{{\rm{CM}}}}} \right)_{{\rm{initial}}}}\\d = \left( {1.727\;{\rm{m}}} \right) - \left( {1.27\;{\rm{m}}} \right)\\d = 0.45\;{\rm{m}}\end{array}\)

Thus, the distance moved by the boat towards the initial position of the 85-kg person is \(0.45\;{\rm{m}}\).

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Most popular questions from this chapter

You are the design engineer in charge of the crashworthiness of new automobile models. Cars are tested by smashing them into fixed, massive barriers at 45 km/h. A new model of mass 1500 kg takes 0.15 s from the time of impact until it is brought to rest. (a) Calculate the average force exerted on the car by the barrier. (b) Calculate the average deceleration of the car in g’s.

At a hydroelectric power plant, water is directed at high speed against turbine blades on an axle that turns an electric generator. For maximum power generation, should the turbine blades be designed so that the water is brought to a dead stop, or so that the water rebounds?

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(c) with bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm. [Hint: The average net force on him, which is related to impulse, is the vector sum of gravity and the force exerted by the ground. See Fig. 7–34.] We will see in Chapter 9 that the force in (b) exceeds the ultimate strength of bone (Table 9–2).

FIGURE 7-34 Problem 24.

A 0.060-kg tennis ball, moving with a speed of 5.50 m/s has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.00 m/s. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision?

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