mechanics

tingyuau · **Joined:** Wed, 5 Oct 2011 21:02:24 UTC **Posts:** 25

A particle of mass m is projected vertically upward with speed u and when it reaches its greatest height a second particle, of mass 2m, is projected vertically upward with speed 2u from the same point as the first. Prove that the time which elapses between the projection of the second particle and its collision with the first is u/4g, and find the height above the point of projection at which the collision occurs. (Ans: (15u^2)/32g ) If, on collision, the particles coalesce prove that the combined particle will reach a greatest height of (19u^2)/18g above the point of projection.

I did it for almost ten times but still get t=2u/g
Can anyone help me???
THANKS!!!

Shadow · **Posted:** Tue, 14 Feb 2012 22:32:00 UTC

tingyuau wrote:

A particle of mass m is projected vertically upward with speed u and when it reaches its greatest height a second particle, of mass 2m, is projected vertically upward with speed 2u from the same point as the first. Prove that the time which elapses between the projection of the second particle and its collision with the first is u/4g, and find the height above the point of projection at which the collision occurs. (Ans: (15u^2)/32g ) If, on collision, the particles coalesce prove that the combined particle will reach a greatest height of (19u^2)/18g above the point of projection.

I did it for almost ten times but still get t=2u/g
Can anyone help me???
THANKS!!!

How did you get that answer?

tingyuau · **Joined:** Wed, 5 Oct 2011 21:02:24 UTC **Posts:** 25

Shadow wrote:

tingyuau wrote:

A particle of mass m is projected vertically upward with speed u and when it reaches its greatest height a second particle, of mass 2m, is projected vertically upward with speed 2u from the same point as the first. Prove that the time which elapses between the projection of the second particle and its collision with the first is u/4g, and find the height above the point of projection at which the collision occurs. (Ans: (15u^2)/32g ) If, on collision, the particles coalesce prove that the combined particle will reach a greatest height of (19u^2)/18g above the point of projection.

I did it for almost ten times but still get t=2u/g
Can anyone help me???
THANKS!!!

How did you get that answer?

I let the greatest height to be h. Then from the first particle, h=ut-1/2gt^2. After it reaches the greatest height, it falls and collides with the second particle. The distance the first particle traveled will be h-y and the distance the second particle travelled will be y so that the total distance equal to h. Then I came up with two equations of motions which are h-y=-1/2ut^2 and y=2ut -1/2gt^2. This is how I got t=2u/g. but it clearly is not the right answer...

Shadow · **Posted:** Wed, 15 Feb 2012 01:09:46 UTC

tingyuau wrote:

Shadow wrote:

tingyuau wrote:

A particle of mass m is projected vertically upward with speed u and when it reaches its greatest height a second particle, of mass 2m, is projected vertically upward with speed 2u from the same point as the first. Prove that the time which elapses between the projection of the second particle and its collision with the first is u/4g, and find the height above the point of projection at which the collision occurs. (Ans: (15u^2)/32g ) If, on collision, the particles coalesce prove that the combined particle will reach a greatest height of (19u^2)/18g above the point of projection.

I did it for almost ten times but still get t=2u/g
Can anyone help me???
THANKS!!!

How did you get that answer?

I let the greatest height to be h. Then from the first particle, h=ut-1/2gt^2. After it reaches the greatest height, it falls and collides with the second particle. The distance the first particle traveled will be h-y and the distance the second particle travelled will be y so that the total distance equal to h. Then I came up with two equations of motions which are h-y=-1/2ut^2 and y=2ut -1/2gt^2. This is how I got t=2u/g. but it clearly is not the right answer...

Let the launch point be height 0. Then we know that u-gt_0=0

after time t_0

when we are at our highest point, i.e. $t_0={u\over g}$ . So $$h=ut_0-{1\over 2}gt_0^2={u^2\over g}-{1\over 2}g\left({u\over g}\right)^2={u^2\over 2g}$ .

Now, at this point the top mass starts falling and at time t-t_0=t'

has velocity $-{1\over 2}gt'^2$ and the other mass has velocity $2u-{1\over 2}gt'^2$ , so they have a relative velocity of

, so since the initial separation distance is ${u^2\over 2g}$ and they have a constant relative velocity of

, we see that the

at which they collide is $$\left({u^2\over 2g}\right)/(2u)={u\over 4g}$ as desired.

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