On a horizontal frictionless frozen lake,a gi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Since there is no external horizontal force acting on the system,the position of the centre of mass of the system remains unchanged.Let the initia

Vedclass · Accepted Answer

$4 \,m$

Vedclass · Answer

(D) Since there is no external horizontal force acting on the system,the position of the centre of mass of the system remains unchanged.Let the initial position of the girl be at the origin $(x = 0)$ and the box be at $x = 20 \,m$.The mass of the girl is $m_1 = 36 \,kg$ and the mass of the box is $m_2 = 9 \,kg$.The initial position of the centre of mass $(X_{CM})$ is:$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{36 	imes 0 + 9 	imes 20}{36 + 9} = \frac{180}{45} = 4 \,m$.Let the girl move a distance $x$ towards the box and the box move a distance $(20 - x)$ towards the girl. When they meet,they will both be at the position of the centre of mass,$X_{CM} = 4 \,m$.Therefore,the girl has travelled $4 \,m$ from her initial position to reach the centre of mass,and the box has travelled $20 - 4 = 16 \,m$ to reach the same point.Thus,the girl has travelled $4 \,m$.

On a horizontal frictionless frozen lake,a girl of mass $36 \,kg$ and a box of mass $9 \,kg$ are connected to each other by means of a rope. Initially,they are $20 \,m$ apart. The girl exerts a horizontal force on the box,pulling it towards her. How far has the girl travelled when she meets the box?

Similar Questions

Which of the following statements regarding the centre of mass is $NOT$ true?

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

$A$ slender uniform rod of length $L$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right,the position of its centre of mass at the time when the rod becomes horizontal is:

The distance between the carbon and oxygen atoms in a carbon monoxide molecule is $1.1 \, \mathring A$. Given that the mass of the carbon atom is $12 \, a.m.u.$ and the mass of the oxygen atom is $16 \, a.m.u.$,find the position of the center of mass of the carbon monoxide molecule.

Vedclass Test Series

Exam Paper Generator

Online Exam Module