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SRIKSHADY - Determined & self-motivated for IIT

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9 Feb 2009 00:09:47 IST

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Two skaters A & B of mass M and 2M are standing together on a frictionless ice surface. They push ea

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Two skaters A & B of mass M and 2M are standing together on a frictionless ice surface. They push each other apart. The skater B moves away from A with a speed of 2 m/s relative to ice. The separation between the two skaters after 5 seconds will be

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Love Goyal

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Joined: 30 May 2008

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9 Feb 2009 00:12:23 IST

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15 m, as there is no xternal force, net momentum will be same(i.e. 0). there for velocity of B is 1 m/s. calculate the dist after 5 sec, easy.plz rate

SRIKSHADY - Determined & self-motivated for IIT

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9 Feb 2009 00:16:14 IST

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Sorry dude, but dats the wrong answer. the options are

1.	30 m

	2.	32 m

	3.	40 m

	4.	28 m

Anubhav

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Joined: 4 Feb 2009

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9 Feb 2009 01:15:03 IST

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Anubhav

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9 Feb 2009 01:27:33 IST

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The ans is 30m. . . . . . How to solve. . . Since initially the velocity of centre of mass was zero, so finally also it will be 0. Therefore final velocity of A is 4m/s in dir. Opp. to that of B. So by using kinematics we can find the distance between them after 5s, . . 30m. CHEERS. . . DONT FORGET TO RATE ME.

Ashutosh Sharma

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9 Feb 2009 11:13:13 IST

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Our system : " skater A + skater B "

No NET external force on our system. so no acceleration. And hence initial vel = final vel =0

So,

Hence, v = -4 ms^-1

Now, Relative velocity of seperation = ( 2 + 4) ms^-1

Hence, seperation in 5 seconds is 6 x 5 = 30m

shraman asa

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10 Feb 2009 16:16:06 IST

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yep all above ar correct .............just use CONSERVATION OF MOMENTUM!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Sai Karthik

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11 Feb 2009 21:14:15 IST

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just conserrv momentum the ans is 4*5 +2*5 = 30

Rohit Khattar

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Joined: 10 Jan 2009

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12 Feb 2009 02:13:47 IST

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hey this was the 1st quesn of AIEEE goiit mock

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