On a smooth horizontal plane,a uniform string | vedclass

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

Question

(B) Let the mass of the man be $M_m = M$ and the mass of the string be $M_s = M$. The length of the string is $L$.Since the plane is smooth,there is n

Vedclass · Accepted Answer

$L / 4$

Vedclass · Answer

(B) Let the mass of the man be $M_m = M$ and the mass of the string be $M_s = M$. The length of the string is $L$.Since the plane is smooth,there is no external horizontal force acting on the system (man + string). Therefore,the position of the center of mass of the system remains unchanged.Let the initial position of the man be $x_m = 0$ and the center of mass of the string be at $x_s = L/2$.The initial position of the center of mass of the system is:$X_{cm} = \frac{M_m x_m + M_s x_s}{M_m + M_s} = \frac{M(0) + M(L/2)}{M + M} = \frac{ML/2}{2M} = L/4$.After the man collects the entire string,the man and the string become a single system of mass $2M$ located at the final position of the man,$x_f$.The final position of the center of mass of the system is $X'_{cm} = x_f$.Since the center of mass does not move,$X_{cm} = X'_{cm}$,so $x_f = L/4$.The displacement of the man is $\Delta x = |x_f - x_m| = |L/4 - 0| = L/4$.

Similar Questions

Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$,by what distance should the second particle be moved,so as to keep the centre of mass at the same position?

Explain translational motion by the given illustration.

Vedclass Test Series

Exam Paper Generator

Online Exam Module