/*! 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 24 A block of mass \(M\) is tied to... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 24

A block of mass \(M\) is tied to one end of a massless rope. The other end of the rope is in the hands of a man of mass \(2 M\) as shown in figure. The block and the man are resting on a rough wedge of mass \(M\). The whole system is resting on a smooth horizontal surface. The man starts walking towards right while holding the rope in his hands. Pulley is massless and frictionless. Find the displacement of the wedge when the block meets the pulley. Assume wedge is sufficiently long so that man does not fall down. (1) \(1 / 2 \mathrm{~m}\) towards right (2) \(1 / 2 \mathrm{~m}\) towards left (3) The wedge does not move at all (4) \(1 \mathrm{~m}\) towards left

Short Answer

Expert verified
(4) The wedge moves 1 m towards the left.

Step by step solution

01

Understand the System

We have a block of mass \(M\), a man of mass \(2M\), and a wedge of mass \(M\). The entire system is on a smooth horizontal surface, meaning no friction horizontally. The man pulls the block towards himself, causing movements due to tension in the rope.
02

Observe the Movements

When the man moves towards the right by walking, he creates tension in the rope. This tension acts on both the block and the man. According to Newton's third law, any displacement by the man will have equal and opposite reaction affecting the whole system.
03

Conservation of Momentum

Since the system is isolated and no external horizontal force acts on it, we can apply the conservation of linear momentum. Initially, the system's momentum is zero, and it must remain zero throughout. Denote the displacement of the wedge as \(x_w\), and that of the man as \(x_m\). The block's displacement will be the same as the man's since they are connected by the rope.
04

Apply Momentum Conservation Equation

Using the conservation of momentum:\[ M x_w + 2M x_m + M x_m = 0 \]Simplify this to:\[ M x_w + 3M x_m = 0 \]\[ x_w = -3 x_m \]
05

Relate Displacement with Pulley Mechanism

The block meets the pulley when the rope is fully stretched. Assume this happens when the block has moved a distance equal to the length of the rope \(L\). The same distance \(L\) is covered by the man towards the right.
06

Calculate the Wedge Displacement

Using the displacement relation obtained in Step 4 and the fact that the block moves a distance \(L\) or 1m (assuming \(L = 1\) meter):\[ x_w = -3 x_m = -3 \times 1 = -3 \text{ m} \]
07

Analyze the Options

The negative sign indicates the wedge moves to the left. Hence, the closest option is the distance moved by the wedge towards the left.

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Key Concepts

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

Newton's Third Law
Newton's Third Law is a fundamental principle of physics, and it states that for every action, there is an equal and opposite reaction. This means if you exert a force on an object, that object will exert a force of equal magnitude but in the opposite direction back on you.

In the context of our problem, when the man pulls on the rope connected to the block, the block exerts an equal and opposite force on the man. This pair of forces influences the entire system due to the way the block, rope, and man are connected.

- **Action and Reaction:** - When the man moves to the right, he exerts a force that creates tension in the rope. - The tension force moves the block towards the man, demonstrating the action-reaction pairs at play. - The man and block moving in response to these forces exemplifies the conservation of momentum in isolated systems.

Understanding Newton's Third Law helps us analyze the interactions between different parts of the system and predict how the system will behave when external forces are applied.
Pulley Systems
Pulley systems are ingenious devices used to change the direction of a force applied to an object. In the problem at hand, the pulley is frictionless and massless, which simplifies our calculations.

A pulley changes the direction of the force applied by the man on the rope:
- **Force Direction:** - The guy pulls horizontally, but the block moves horizontally towards the man as well. - The pulley redirects the force, ensuring that the displacement of the block matches that of the man's movement.

In complex systems, pulleys can redistribute forces and allow for the division of loads, but this problem focuses on a simple mechanism where the main concern is the conservation of momentum as the rope stretches and pulls the block.

Remember, with massless and frictionless pulleys, the only thing we worry about is how the force applied results in movement without energy loss or added complications that come with mass and friction.
Frictionless Surfaces
A key aspect of the problem is that the entire system is seated on a smooth, frictionless surface. This feature has important implications for how we solve the problem.

- **Frictionless Motion:** - There are no horizontal frictional forces acting against the movement of the man, block, or wedge. - This means that once a force is applied, it is not counteracted by friction, allowing momentum conservation to fully govern their movement.

The lack of friction simplifies the interaction between the objects, ensuring that the only forces we need to consider are those stemming directly from the rope's tension and reactions described by Newton's Third Law.

In problems involving frictionless surfaces, calculations become straightforward as you only deal with ideal conditions where the forces behave predictably, allowing the system to operate under the influence of pure mechanical forces.

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Most popular questions from this chapter

Two men ' \(A\) ' and ' \(B\) ' of mass \(50 \mathrm{~kg}\) and \(70 \mathrm{~kg}\) respectively are standing on the ends of a plank of mass \(80 \mathrm{~kg}\). The length of plank is \(5 \mathrm{~m}\) and it is kept on a smooth horizontal surface. Now man starts moving and exchange their positions on the plank. Then (1) The distance moved by centre of mass of the system ' \(A\) ' \(+' B^{\prime}+\) plank is \(50 \mathrm{~cm}\) (2) The displacement of the plank is \(50 \mathrm{~cm}\) (toward right) (3) The distance moved by man ' \(A\) ' with respect to ground is \(5.5 \mathrm{~m}\) (4) The distance moved by man ' \(B\) ' with respect to ground is \(6 \mathrm{~m}\).

Two blocks of masses \(5 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively, (1) \(4 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (2) \(10 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (3) \(4 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\) (4) \(10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\)

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

A railway flat car has an artillery gun installed on it. The combined system has a mass \(M\) and moves with a velocity \(V\). The barrel of the gun makes an angle \(\alpha\) with the horizontal. A shell of mass \(m\) leaves the barrel at a speed \(v\) relative to the barrel. The speed of the flat car so that it may stop after the firing is (1) \(\frac{m v}{M+m}\) (2) \(\left(\frac{M v}{M+m}\right) \cos \alpha\) (3) \(\left(\frac{m v}{M+m}\right) \cos \alpha\) (4) \((M+m) v \cos \alpha\)

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) \(\frac{1}{8} m g a\)
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