Ram & Shyam are on a boat for a picnic. The boat is initially at rest. Ram has a big watermelon which he throws towards shyam. Shyam catches the melon and eats half of it. He throws back the remaining

Question

Ram & Shyam are on a boat for a picnic. The boat is initially at rest. Ram has a big watermelon which he throws towards shyam. Shyam catches the melon and eats half of it. He throws back the remaining half to Ram. He eats half of the melon that he receives & throws the remaining part to Shyam. Shyam again eats half of what he receives and returns the remaining part to Ram. This continues till the melon lasts. The two are sitting at the two ends of the boat which has length L. Combined mass of the boat, Ram & Shyam is $M_{0}$ and the mass of the water melon is M. Assume that the boat can move horizontally on water without any resistance. What is the displacement of the boat when watermelon gets finished:-

Options:

Answer:

D

Solution:

Shyam eats $\frac{M}{2} + \frac{M}{8} + \frac{M}{32} + \dots = \frac{M /2}{1 - 1/4} = \frac{2 M}{3}$ ; Ram keeps $\frac{M}{3}$ .

Net effect: mass $\frac{2 M}{3}$ shifted from Ram's end to Shyam's end across $L$ .

No external force ⇒ COM of boat + men + melon stays fixed: $x_{b o a t} = \frac{( 2 M /3 ) L}{M _{0} + M} = \frac{2 M L}{3 ( M _{0} + M )}$ - (D)

TestHub · Accepted Answer

Shyam eats 2 M ​ + 8 M ​ + 32 M ​ + … = 1 − 1/4 M /2 ​ = 3 2 M ​ ; Ram keeps 3 M ​ . Net effect: mass 3 2 M ​ shifted from Ram's end to Shyam's end across L . No external force ⇒ COM of boat + men + melon stays fixed: x b o a t ​ = M 0 ​ + M ( 2 M /3 ) L ​ = 3 ( M 0 ​ + M ) 2 M L ​ - (D)

Physics - Center of Mass Question with Solution | TestHub

Options:

Doubts & Discussion