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Mechanics

Blazing goIITian

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16 Sep 2007 20:36:38 IST

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808

newtons laws of motion 1

None

a car starts frm rest and accelerates uniformly with 2 m/s² . At t=10s a stone is dropped out of the window(1m high) of the car.

find the
A)velocty

B)acceleration of the stone at 10.1 sec.

neglect air resistance and take g = 9.8 m/s^{2

guys please tell me only which concept(s) to apply.
no need of givcing me the ans.

cheers.....!!!!}

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Comments (8)

Ankur Limaye

Cool goIITian

Joined: 7 Jul 2007

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16 Sep 2007 21:18:03 IST

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This seems easy .....
The ball is thrown outside the window ......... this means it will have the horizontal velocity that of the car at that instant ....... and no vertical velocity initially .....
Then, there even collisions can occur as the time is too large ......

I maybe wrong ...... just tried it .......

Swashata Ghosh

Blazing goIITian

Joined: 20 May 2007

Posts: 459

16 Sep 2007 21:23:21 IST

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Well sinjan this is an easy one.

1. First find out the velocity of the car at t=10 sec using the formula

v = u + at [Here u = 0]

    Due to inertia this velocity will be the velocity of the stone.

2. When the stone is projected from the car then it only experiences an acc. due to gravity ie g. So no matter what be the time the acc. of the stone will be g.
    If after 10.1 sec from projection the stone reaches the ground then the wt of the stone will be balannced by the normal rxn provided by the ground on the stone and net acc will be 0, otherwise it will fall freely with an acc of g.
    I think the question is on finding the velocity of the stone after 10.1 sec of projection. If so then find out the vertical velocity of the stone after 10.1 sec(say z), and also check whether it will reach the ground or not. If it reaches the ground then the velocity will be 0. If not then the resultant velocity will be the resulant of the vectors v and z. Also find out the angle of the resultant velocity with using the formula of the vector...

    These are the concepts. If u want the soln. then nudge me!!!!!!!!!!!!!

Ankur Limaye

Cool goIITian

Joined: 7 Jul 2007

Posts: 98

16 Sep 2007 21:45:08 IST

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Ya, the magnitude of accleration will be always be g .......

But, as the time period is too long ........ i think the ball will keep bouncing ...... not to the same height ( as it is not perfectly elastic collision) ...... but surely it is not perfectly inelastic collision ......

Swashata Ghosh

Blazing goIITian

Joined: 20 May 2007

Posts: 459

16 Sep 2007 21:50:43 IST

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In that case the co-eff of the restitution should also be given!!!!!!!!

Ankur Limaye

Cool goIITian

Joined: 7 Jul 2007

Posts: 98

16 Sep 2007 22:02:14 IST

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Yup ......
The question is not complete .......

Pramit

Cool goIITian

Joined: 11 Apr 2007

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16 Sep 2007 23:34:54 IST

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Its so simple yaaar.......
At 10.1 s the ball wiil have horizontal velocity of car at t=10s and a vertical velocity downward g/10 which it has get in 0.1 s
and off course its acceleration will be g with respect to earth and square root of (g sq. + a sq. ) with respect to car

and dontforget to rate me

James Bond

Hot goIITian

Joined: 14 Sep 2007

Posts: 188

21 Sep 2007 17:36:44 IST

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Here's how it looks...

Nudge me if more req...

Rate me...

Swashata Ghosh

Blazing goIITian

Joined: 20 May 2007

Posts: 459

21 Sep 2007 17:46:22 IST

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Good pic 12121212..... But it takes a lot of time.. isn't it!!!!!!!!!

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