AIEEE - rotational mechanics

A man of mass 'm' stands on a platform of mass 'M' kept on a ice block (friction = 0). If the man walks on the platform with a velocity 'v' w.r.t the platform, the relative velocity of the platform, the relative velocity of the platform w.r.t the ice block is

(a) \frac{mv}{M-m}

(b) \frac{M-m}{mv}

(c) \frac{Mv}{M+m}

(d) \frac{M+m}{Mv}

13 Answers

24
eureka123 ·

m(v'+v)+Mv'=0 (i have ignored direction analysis)
=>mv'+mv+Mv'=0
=>v'=-mv/m+M

11
Anirudh Narayanan ·

if u see carefully, u'll notice that there's no such option

1
°ღ•๓яυΠ·

i gues sthis is a soolvd eg in hcv just check ..... page 153 q 15 .if ma memroy wrks well....

1
kartik sai ·

IT IS VERY SIMPLE THEN......YOUR OPTIONS R WRONG !!!

11
Anirudh Narayanan ·

11
Anirudh Narayanan ·

someone, pls reply

1
JOHNCENA IS BACK ·

(a)1ms-2

39
Dr.House ·

39
Dr.House ·

use 1 and 2 u should get

11
Anirudh Narayanan ·

thanx b555

11
Anirudh Narayanan ·

just one doubt: isn't tension in a string same throughout the string???

106
Asish Mahapatra ·

@Ani: If the tensions are same then the pulley will not rotate. But in this case the pulley rotates and hence T on both sides MUST be different naa to cause the rotation motion of the pulley

11
Anirudh Narayanan ·

oh...ok ...thanx asish

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