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Mechanics

Cool goIITian

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7 Jan 2008 21:31:39 IST

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585

Very Logical Question..Plz explain the concept!!

None

Here, in the figure, we see 6 balls of similar masses attached.
Now, if we lift two balls from the left hand side, and leave them to fall down to strike the remaining balls, two balls from the right will lift up.

I'd like to know why always 2 balls would lift up?
Why cant one single ball lift up with an equal momentum, i.e.m(2v) and not (mv + mv).

Plz explain this...

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Comments (9)

abc def

Hot goIITian

Joined: 27 Oct 2007

Posts: 102

7 Jan 2008 22:11:01 IST

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Answer is simple. I have assumed that you lifted both the balls and they were released after a time interval. The first ball strikes causing the last ball to rise , then the second one strikes casusing the second - last to rise up. Note that this will only happen if the second ball hits the first one just just & just after it collides with the first. At one point , the velocity of the first ball during collision is zero. Thus it is like lifting one ball and striking. Use collisions equations and brood at the magnitude of the force "travelled" to the last ball during the course of all the collisions. Please note that the balls have to fight gravity to rise up , so the oribablity of two shots in one hit is not so possible (Especially if the BALL IS RELEASED AND NOT THROWN WITH A INITIAL VELOCLTY).

Hope it helps. Rate if useful.

Saurabh Manchanda

Cool goIITian

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8 Jan 2008 21:23:03 IST

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I'm speaking exactly opposite of ur assumption..

I'm not releasing the 2 balls after a time interval.. I'm releasing the balls simultaneously..
Then even two balls will rise with momentum ( mv + mv)
and not 1 ball with a momentum ( m *2v)
Why???

Rohith B.V

Blazing goIITian

Joined: 16 May 2007

Posts: 427

8 Jan 2008 22:00:01 IST

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good doubt goiit_user!!! i think this is the answer....since no external force acts on the newtons cradle...the momentum equals initial momentum...this is right...BUT...since no net external force acts on the system during this period....KE must be conserved...if u calculate the initial KE it comes out to be mv² so final ke must also be mv² so we get that mv²=(mv_1²+mv_2²)/2 -----1 and
2mv=mv₁+mv₂ -----2

these two equations can be simultaneously satisfied ONLY if each of the balls gets a velocity v...hence both the balls must rise and the momentum is distributed as mv+mv...

Priyesh

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8 Jan 2008 22:00:51 IST

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assuming elastic collision

kinetic energy should also be conserved so initial K.E is mv^2

for ur case : if one ball lifts with speed 2v then K.E becomes 1/2m*4v^2 = 2mv^2

so u see that K.E is not conserved in ur case hence two balls will lift

Priyesh

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8 Jan 2008 22:02:14 IST

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oh just posted before by rohith

Saurabh Manchanda

Cool goIITian

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8 Jan 2008 22:13:06 IST

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Thx everbody...

What would happen if the ball on the righmost side is of mass 2m...Then, will 2 balls lift or only the last ball ??

Gaurav |spideyunlimited| Ragtah

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8 Jan 2008 22:18:22 IST

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2 basic concepts are used in this apparatus which is called NEWTON's CRADLE.
1st) conservation of momentum
2nd) Centre of mass remains at rest in this setup.

Priyesh

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8 Jan 2008 22:28:15 IST

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make the equations

let first ball of rightmost side 2m mass be lifted with speed v1 & the ball next to rightmost (of mass m) with v2

now conserving energy

mv^2 = mv1^2 + (mv2^2)/2

conserving momentum

2mv = 2mv1 + mv2

from these two equations v1v2 = v2^2 / 4

=. either v2 = 0 or v2 = 4v1

Gaurav |spideyunlimited| Ragtah

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8 Jan 2008 23:00:20 IST

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V2 wont be zero... if its zero then centre of mass will not remain stationary so v2 = 4v1 is the answer

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