/*! 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 57 A man of mass \(m\) stands on a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 57

A man of mass \(m\) stands on a ladder which is tied to a free balloon of mass \(M\). The balloon is at rest initially. If the man starts to climb the ladder at a constant velocity \(v\) relative to the ladder, then initial speed of balloon will be (neglect mass of ladder) (A) \(\frac{m v}{M+m}\) (B) \(\frac{m v}{M+2 m}\) (C) \(\frac{M v}{M+m}\) (D) \(\frac{m v}{M}\)

Short Answer

Expert verified
The initial speed of the balloon will be \(\textbf{(D)}\) \(V' = \frac{m v}{M}\).

Step by step solution

01

Identify the initial momentum of the system.

Before the man starts climbing the ladder, both the man and the balloon are at rest. Therefore, their initial combined momentum is zero.
02

Write the momentum conservation equation.

According to the conservation of momentum principle, the total momentum before and after the man starts climbing must be equal. We can express this mathematically as: \(m_{1} v_{1} + m_{2} v_{2} = m_{1} v'_{1} + m_{2} v'_{2}\) where \(m_{1}\) and \(m_{2}\) are the masses of the man and balloon, respectively; \(v_{1}\) and \(v_{2}\) are their initial velocities (both 0 in this case); and \(v'_{1}\) and \(v'_{2}\) are their final velocities when the man is climbing the ladder at velocity \(v\).
03

Substitute given values into the equation.

Now, we can substitute the given values into the equation: \(m(0) + M(0) = m(v) + M(V')\)
04

Solve for the final velocity of the balloon.

Our goal is to find the initial speed (or final velocity) of the balloon, \(V'\). We can rearrange the equation to solve for \(V'\): \(M(V') = m(v)\) \(V' = \frac{m(v)}{M}\)
05

Match the result with the answer choices.

Comparing our result with the answer choices, we can see that the initial speed of the balloon is: \(\textbf{(D) } V' = \frac{m v}{M}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle of mass \(m_{1}\) moving with velocity \(u_{1}\) strikes another particle of mass \(m_{2}\) moving with velocity \(u_{2}\). After collision the velocities of the particles are \(v_{1}\) and \(v_{2}\) respectively. Both the particles are moving on a frictionless horizontal surface. Column-I represents the nature of collision between the particles and Column-II represents the equations for physical quantities. (Symbols have their usual meanings) Column-I (A) Head on elastic collision (B) Head on inelastic collision (C) Oblique elastic collision (D) Oblique inelastic collision Column-II 1\. \(m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}\) \(=m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}\) 2\. \(e=\frac{\left|\vec{v}_{2}\right|+\left|\overrightarrow{v_{1}}\right|}{\left|\overrightarrow{u_{1}}\right|+\left|\overrightarrow{u_{2}}\right|}\) \(\begin{aligned} \text { 3. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\ &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\\ \text { 4. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\\ &>\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \end{aligned}\)

Two blocks \(A\) and \(B\) of mass \(m\) and \(2 \mathrm{~m}\) respectively are connected by a massless spring of force constant \(k\). They are placed on a smooth horizontal plane. They are stretched by an amount \(x\) and then released. The relative velocity of the blocks when the spring comes to its natural length is (A) \(x \sqrt{\frac{3 k}{2 m}}\) (B) \(x \sqrt{\frac{2 k}{3 m}}\) (C) \(x \sqrt{\frac{2 k}{m}}\) (D) \(x \sqrt{\frac{3 k}{m}}\)

Consider a block \(P\) of mass \(2 M\) placed on another block \(Q\) of mass \(4 M\) lying on a fixed rough horizontal surface of coefficient of friction \(\mu\) between this surface and the block \(Q\). Surfaces of \(P\) and \(Q\) interacting with each other are smooth. \(A\) mass \(M\) moving in the horizontal direction along a line passing through the centre of mass of block \(Q\) is normal to the face. Speed of mass \(M\) is \(v_{0}\) when it collides elastically with the block \(Q\) at a height \(s\) from the rough surface. Both the blocks have same length \(4 s\). Velocity \(v_{0}\) is enough to make the block \(P\) topple. When mass \(M\) collides with block \(Q\) with velocity \(v_{0}\) the block \(P\). (A) Does not move (B) Moves forward (C) Moves backward (D) Either (B) or (C)

Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2: Principle of conservation of momentum holds true for all kinds of collisions. (A) Statement-1 is true, Statement- 2 is true; Statement- 2 is the correct explanation of Statement- 1 (B) Statement-1 is true, Statement- 2 is true; Statement- 2 is not the correct explanation of Statement-1 (C) Statement- 1 is false, Statement- 2 is true. (D) Statement- 1 is true, Statement- 2 is false.

Two skaters of masses \(40 \mathrm{~kg}\) and \(60 \mathrm{~kg}\) respectively stand facing each other at \(S_{1}\) and \(S_{2}\) where \(S_{1} S_{2}\) is \(5 \mathrm{~m}\). They pull on a massless rope stretched them, then they meet at (A) \(2.5 \mathrm{~m}\) from \(S_{1}\) and \(S_{2}\) (B) \(3 \mathrm{~m}\) from \(S_{1}\) and \(2 \mathrm{~m}\) from \(S_{2}\) (C) \(2 \mathrm{~m}\) from \(S_{1}\) and \(3 \mathrm{~m}\) from \(S_{2}\) (D) \(3 \mathrm{~m}\) from \(S_{1}\) and \(8 \mathrm{~m}\) from \(S_{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Radiation

Read Explanation

Electrostatics

Read Explanation

Physics of Motion

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.