Question

A cart of mass M is tied to one end of a massless rope of length 10m. The other end of the rope is in the hands of a man of mass M/2. The entire system is on a smooth horizontal surface. If the man pulls the cart by the rope then find the
distance by which the man will slip on the horizontal surface before the man and the cart will meet each other.

(1) 20/3m
(2) 10/3m
(3) 0
(4) None​

Answer 1

answer : option (1) 20/3 m

explanation : external force acting on the system of man and cart , $F_{ext}=0$

⇒$a_{C.M}=\frac{dv_{C.M}}{dt}=0$

⇒$v_{C.M}=0$

so, centre of mass of the system remains same.

let position of man is (0,0) and cart is (10,0).

so, position of man with respect to cart , $X_{C.M}$ = (M/2 × 0 + M × 10)/(M/2 + M)

= 20/3 m

so, 20/3 m distance by which the man will slip on the horizontal surface before the man and the cart will meet each other.

Answer 2

$\huge\bold\purple{Answer:-}$

(1) 20/3m