/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A man stands at one end of a boa... [FREE SOLUTION] | 91Ó°ÊÓ

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A man stands at one end of a boat which is stationary in water. Neglect water resistance. The man now moves to the other end of the boat and again becomes stationary. The centre of mass of the 'man plus boat' system will remain stationary with respect to water (1) only when the man is stationary initially and finally (2) only if the man moves without acceleration on the boat (3) only if the man and the boat have equal masses (4) in all cases

Short Answer

Expert verified
The center of mass remains stationary in all cases, answer is option (4).

Step by step solution

01

Understanding the Problem

The problem involves a man moving from one end of a stationary boat to the other, and we need to determine under what conditions the center of mass of the entire 'man plus boat' system remains stationary with respect to the water.
02

Defining the System

The system in question consists of two parts: the man and the boat. We know from physics that the center of mass of a closed system is unaffected by internal forces and remains stationary if there are no external forces acting on it.
03

Analyzing the Center of Mass

The center of mass of the 'man plus boat' system will remain stationary since there are no external forces acting (negligible water resistance). The movement of the man is an internal force within the system.
04

Exploring the Options

Let's consider the options: 1. The man being stationary initially and finally does not affect the internal forces acting. 2. Movement with or without acceleration does not affect the center of mass, as it concerns only internal motion. 3. Equal masses do not influence the conditions of internal forces and external forces are absent. 4. Since internal forces do not change the center of mass, it remains stationary in all cases.
05

Conclusion

Therefore, the center of mass of the system remains stationary with respect to the water in all cases given there are no external forces acting. This supports option (4).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Forces
In any system, internal forces are those that occur between the objects within the system. These are forces that are generated by the components of the system itself. For our boat scenario, the internal forces include the actions exerted by the man moving from one end of the boat to the other.

Importantly, internal forces do not alter the overall center of mass of the system. This is because each force within the system has an equal and opposite reaction due to Newton's Third Law. As a result, they cancel each other out. The movement of the man within the boat is an example of such internal forces being at play. The man pushes against the boat, and the boat pushes back on him with an equal force.

This characteristic of internal forces is key when analyzing the center of mass in a closed system like the man and boat combination. Because these forces are contained within the system, they do not affect the overall position of the center of mass.
External Forces
External forces are those that influence a system from the outside. These forces could significantly alter the position of the center of mass of a system if they were present. In the exercise of the man and the boat, one of the key points is the absence of significant external forces like water resistance.

Because we are instructed to neglect water resistance, we conclude that no external forces act on the system. Therefore, it's determined that the center of mass remains unaffected by anything outside of the man and boat system. The water does not exert a significant force on the boat as it moves, meaning the only forces at play are internal. This condition is crucial because, without external forces, the center of mass of a closed system remains constant and stable.
System Dynamics
System dynamics involves analyzing how the various parts of a system interact and affect each other's motion. In the case of our man and boat system, the dynamics involve understanding how the internal movements within the system determine the overall behavior.

When the man moves, the boat might seem to shift in the opposite direction. This visual change is due to the internal interactions within the system. The man creates momentum in one direction, and to conserve momentum (a principle we'll discuss later), the boat shifts slightly in the opposite direction.

This reaction might create an illusion of movement, but it actually highlights how internal forces manage to maintain the overall balance of the system even when individual parts are in motion. The dynamics of the system ensure that despite individual movements, the system remains in equilibrium in terms of its center of mass.
Conservation of Momentum
Conservation of momentum is a pivotal principle in physics, dictating that in the absence of external forces, the momentum of a system remains constant over time. When the man moves across the boat, this principle plays a crucial role in ensuring the center of mass remains stationary.

As the man steps forward, he exerts force on the boat. According to the conservation of momentum, the boat moves slightly in the opposite direction to maintain the overall momentum of the system. This is what keeps the center of mass in the same position relative to the external environment.

By maintaining the total momentum, the system dynamics ensure stability of the center of mass. Thus, even though there seems to be movement within the system, the overall state of neutral momentum conservation implies the system’s center of mass remains unchanged relative to the water.

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Most popular questions from this chapter

An object initially at rest explodes into three fragments \(A\) \(B\) and \(C\). The momentum of \(A\) is \(p \hat{i}\) and that of \(B\) is \(\sqrt{3} p]\) where \(p\) is a \(+\) ve number. The momentum of \(C\) will be (1) \((1+\sqrt{3}) p\) in a direction making angle \(120^{\circ}\) with that 0 \(A\) (2) \((1+\sqrt{3}) p\) in a direction making angle \(150^{\circ}\) with that \(d\) \(B\) (3) \(2 p\) in a direction making angle \(150^{\circ}\) with that of \(A\) (4) \(2 p\) in a direction making angle \(150^{\circ}\) with that of \(B\)

For a two-body system in absence of external forces, the kinetic energy as measured from ground frame is \(K_{0}\) and from centre of mass frame is \(K_{\mathrm{cm}}\). Pick up the correct statements. (1) The kinetic energy as measured from centre of mass frame is least. (2) Only the portion of energy \(K_{\mathrm{cm}}\) can be transformed from one form to another due to internal changes in the system. (3) The system always retains at least \(K_{0}-K_{\mathrm{cm}}\) amount of kinetic energy as measured from ground frame irrespective of any kind of internal changes in the system. (4) The system always retains at least \(K_{\mathrm{cm}}\) amount of kinetie energy as measured from ground frame irrespection energy as measured from ground frame irrespective of any kind of internal changes in the system

Figure shows a hollow cube of side ' \(a\) ' and volume ' \(V\). There is a small chamber of volume \(V / 4\) in the cube as shown. The chamber is completely filled by \(m \mathrm{~kg}\) of water. Water leaks through a hole \(H\) and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is (1) \(\frac{1}{2} m g a\) (2) \(\frac{3}{8} m g a\) (3) \(\frac{5}{8} m g a\) (4) \(\frac{1}{8} m g a\)

A particle of mass \(2 m\) is projected at an angle of \(45^{\circ}\) with the horizontal with a velocity of \(20 \sqrt{2} \mathrm{~m} / \mathrm{s}\). After \(1 \mathrm{~s}\) of explosion, the particle breaks into two equal pieces. As a result of this one part comes to rest. The maximum height from the ground attained by the other part is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) \(50 \mathrm{~m}\) (2) \(25 \mathrm{~m}\) (3) \(40 \mathrm{~m}\) (4) \(35 \mathrm{~m}\)

A circular plate of uniform thickness has a diameter of \(28 \mathrm{~cm}\). A circular portion of diameter \(21 \mathrm{~cm}\) is removed from the plate as shown. \(O\) is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from 'O' by (1) \(5 \mathrm{~cm}\) (2) \(9 \mathrm{~cm}\) (3) \(4.5 \mathrm{~cm}\) (4) \(5.5 \mathrm{~cm}\)

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