Rotational Motion Question 34
Question: A boy of mass 20 kg is standing on a 80 kg free to move long cart. There is negligible friction between cart and ground. Initially, the boy is standing 25 m from a wall. If he walks 10 m on the cart toward the wall, then the final distance of the boy from the wall will be:
[JEE ONLINE 23-04-2013]
Options:
A) 15 m
B) 12.5 m
C) 15.5 m
D) 17 m
Show Answer
Answer:
Correct Answer: D
Solution:
If he walks 10 m on the cart towards the wall, then the final distance of the boy from the wall will be. 17m.
Given:Mass of boy m1=20 kg
Mass of cart m2=80 kg
Initial distance between boy and cart is 25 m.
Distance moved by boy towards wall is 10 m
As there is no external force,
so displacement of centre of mass of the (cart + boy) system parallel to the surface is zero.
∴x cm=m1x1 + m2x2 m1 + m2=0
Let when the boy moves 10 m towards the wall, the cart moves away from the wall a distance xSo, displacement of man w.r.t. ground towards the wall is
x1=10−x
And the displacement of cart w.r.t. ground towards the wall is
x2=−x20 × (10−x)+(80 × (−x)) = 0
⇒x=2 m
i.e. Final distance between boy and wall,=25−10+2=17 m
Hence, option (D) is correct.
