
A balloon having mass $m$ is filled with a gas and is held in the hands of a boy. Then suddenly released and gas starts coming out of it with a constant rate. The velocity of the ejected gases is constant $2m/s$ with respect to the balloon. Find out the velocity of the balloon when the mass is reduced to half.
A) ${\ln2}$
B) $2{\ln4}$
C) $2{\ln2}$
D) None of these
Answer
133.8k+ views
Hint: Recall that momentum is the product of mass and velocity of the bodies or the objects. The external force on a system is directly proportional to the rate of change of momentum. The change in momentum takes place in the direction of applied force.
Complete step by step solution:
Step I: Since the balloon is suddenly released and gas starts leaking out, it is clear that there is no external force applied on the system. This means the external force is zero.
Also $F = \dfrac{{dp}}{{dt}}$
Where $F$is the force
$\dfrac{{dp}}{{dt}}$is the rate of change of momentum
Step II: Since external force on the balloon is zero, so
$\dfrac{{dp}}{{dt}} = 0$
$\dfrac{{d(mv)}}{{dt}} = 0$
Step III:
$v\dfrac{{dm}}{{dt}} + m\dfrac{{dv}}{{dt}} = 0$
$m\dfrac{{dv}}{{dt}} = - v\dfrac{{dm}}{{dt}}$
$mdv = - vdm$
$\dfrac{{dm}}{m} = - \dfrac{{dv}}{v}$
Step IV: When the gas is filled then the mass of the balloon is $M$ and velocity will be $0$.
But when the gas starts leaking the mass is $\dfrac{M}{2}$and velocity will be $M$
Integrating above equation with the given conditions,
$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \int\limits_0^M {\dfrac{{dv}}{v}} $
$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \dfrac{1}{v}\int\limits_0^M {dv} $
The derivative of the final velocity of the object gives its initial velocity that is ‘$u$’.
$\ln \dfrac{m}{{2m}} = - \dfrac{1}{v}(u)$
$u = - v\ln \dfrac{1}{2}$
$u = v\ln 2$
$u = 2\ln 2$
Step V: So when the gas leaks out and the mass of the balloon is reduced to half, then the velocity of the balloon will be $2\ln 2$.
Therefore Option (C) is the right answer.
Note: It is to be noted that any object that has mass and is moving has momentum. But the change of momentum of an object is equal to its impulse. Also impulse is a quantity related to force but it is not equal to momentum. Impulse of a body is the increase or decrease in momentum.
Complete step by step solution:
Step I: Since the balloon is suddenly released and gas starts leaking out, it is clear that there is no external force applied on the system. This means the external force is zero.
Also $F = \dfrac{{dp}}{{dt}}$
Where $F$is the force
$\dfrac{{dp}}{{dt}}$is the rate of change of momentum
Step II: Since external force on the balloon is zero, so
$\dfrac{{dp}}{{dt}} = 0$
$\dfrac{{d(mv)}}{{dt}} = 0$
Step III:
$v\dfrac{{dm}}{{dt}} + m\dfrac{{dv}}{{dt}} = 0$
$m\dfrac{{dv}}{{dt}} = - v\dfrac{{dm}}{{dt}}$
$mdv = - vdm$
$\dfrac{{dm}}{m} = - \dfrac{{dv}}{v}$
Step IV: When the gas is filled then the mass of the balloon is $M$ and velocity will be $0$.
But when the gas starts leaking the mass is $\dfrac{M}{2}$and velocity will be $M$
Integrating above equation with the given conditions,
$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \int\limits_0^M {\dfrac{{dv}}{v}} $
$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \dfrac{1}{v}\int\limits_0^M {dv} $
The derivative of the final velocity of the object gives its initial velocity that is ‘$u$’.
$\ln \dfrac{m}{{2m}} = - \dfrac{1}{v}(u)$
$u = - v\ln \dfrac{1}{2}$
$u = v\ln 2$
$u = 2\ln 2$
Step V: So when the gas leaks out and the mass of the balloon is reduced to half, then the velocity of the balloon will be $2\ln 2$.
Therefore Option (C) is the right answer.
Note: It is to be noted that any object that has mass and is moving has momentum. But the change of momentum of an object is equal to its impulse. Also impulse is a quantity related to force but it is not equal to momentum. Impulse of a body is the increase or decrease in momentum.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
