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Further Maths M1

JBKProductions

asdf

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Reply 1

Adjective

Imagine you have a brand new situation, in which one particle is dropped, from rest, from a height of 22.48m, and another is released vertically upwards from O at 15m/s.

What value (velocity, distance from O, time travelling, etc) would the two particles have in common if they were to collide?

Reply 2

Swayum

Work out an equation for each of their displacements (from the same point). When they collide, their displacements are equal.

Reply 3

JBKProductions

Adje

I can't think of what they would have in common except acceleration

Reply 4

JBKProductions

Swayum

Work out an equation for each of their displacements (from the same point). When they collide, their displacements are equal.

This is confusing how would their displacement be equal? Thanks for helping

Reply 5

Swayum

What does a collision mean?

Spoiler

Reply 6

JBKProductions

Swayum

What does a collision mean?

Spoiler

oh ok

Reply 7

Brook Taylor

They wouldn't necessarily have the same acceleration at all, that is irrelevant to the question. Treat the two particles separately, and form an equation for each which describes their displacement from O in terms of the time traveled. The question asks you to show that they collide 1.5 seconds after the second particle is sent upwards, so all you need to do is show that both particles (bearing in mind the first particle starts at a displacement of 22.48m and has a velocity of 0m/s when the second particle starts moving) have the same displacement at the same time - this is the definition, in this scenario, of a collision.

Reply 8

JBKProductions

Brook Taylor

So do i work out how long it takes the first particle to reach 22.48 and then for the second particle put the time as t = t-(the time is took to get to 22.48) and put S = S so then i can substitute it in?

Reply 9

cheemaj187

s = ut + 0.5at^2

Put values for both into this equation and make them equal.

You have now formed simultaneous equations where the only unknown is t...

Reply 10

Brook Taylor

JBKProductions

No, you can make it simpler than that. Call the time that the second particle starts moving t=0. When t=0, the second particle is at S=0, moving with the given velocity, and the first particle is at position S=22.48, moving with a velocity of 0m/s (this is because the particle has reached the peak of it's arc [yes, it's an arc, even if it has no horizontal displacement]). Bearing in mind that the particles both have the same acceleration (technicalities of differing gravitational forces aside), logic states that these two particles will collide, ie. they will at one point in time have the same displacement S.

Reply 11

JBKProductions

Brook Taylor

If S = 22.48 in the first particle and the second particle its S=0 then how would i substitute it in the equation?

Reply 12

cheemaj187

Provided you are taking their displacements from the same point, when t=1.5 their displacements will be equal.

Reply 13

JBKProductions

cheemaj187

Provided you are taking their displacements from the same point, when t=1.5 their displacements will be equal.

so would the first particle thats drops from 22.48m have t=t+1.5 and the second particle t=1.5 to have the same displacement then?

Reply 14

Brook Taylor

JBKProductions

If S = 22.48 in the first particle and the second particle its S=0 then how would i substitute it in the equation?

As I said before, you are forming two separate equations, one for each particle. I suggest you use the formula cheemaj187 suggested in order to form a pair of simultaneous equations which can be solved to find t. Each equation will be in the form of S=?, and because the two particles collide, their displacement S will be the same, so you can essentially equate the two equations to each other and solve to find the time, t, and afterwards plugging in the calculated value of t to find what the displacement is of the collision.

Edit: And no, as I said before, call t=0 the time when the first particle is at its peak of 22.48m above O, and when the second particle is launched from O. They share the same value of 't' throughout if you do as I suggest.

Reply 15

JBKProductions

Brook Taylor

oh ok its very confusing lol

Reply 16

Brook Taylor

JBKProductions

oh ok its very confusing lol

It can be confusing, I agree. With the majority of Mechanics questions that can't be solved by simple logic and/or arithmetic in your head, I suggest you draw a rough diagram of what's happening in the question. For example, in this question, looking just at part (ii), you could draw a profile view of the ground, the second particle on the ground with an arrow indicating its velocity, and the first particle in the air with its displacement labeled and again, an arrow indicating its velocity (which in this case is 0). That would be the situation in the question at time t=0, ie. when the second particle is launched.

Drawing such a diagram makes it a lot easier to picture the situation, and will help reveal certain facts that you might not have realised otherwise (for example, looking at the diagram, it is obvious that gravity, ie. the acceleration of both particles, is working "downwards", in the opposite direction to the initial movement of the second particle).

Incidentally, if you want any further help with any Maths-related questions, just PM me.

Reply 17

cheemaj187

JBKProductions

so would the first particle thats drops from 22.48m have t=t+1.5 and the second particle t=1.5 to have the same displacement then?

no because you are taking values measured from the same point. When the particles collide they will have the same displacement from point A. So when taking measurements from point A only we know that their displacements are equal at time, t.

This is hard to explain without a diagram but try to ignore that the particles are at different points. For example, when they collide the particle at 22.48m could be Xm from its starting point, point B, and the particle at the lower level could be Ym from its starting point, point A. But they are both Ym from point A so provided that you take measurements as if they both started at A you know their displacements are equal at that time.

Reply 18

JBKProductions

cheemaj187

lol sorry but its confusing would S=S when identifying the displacement, time etc and t=0 then?

Reply 19

Brook Taylor

The OP PM'd me and I replied with the following, just in case anyone else was wondering how to do the question:

When the two particles collide, their 't's and 'S's will be equal. We'll look at each particle individually, forming an equation in each in terms of 't', and equating them.

We'll use the following formula: s = ut + 0.5at^2. Because a=-9.8 throughout (this is negative because we're taking upwards velocities as positive), this can be simplified to: s = ut - 4.9t^2

Looking at the first particle, u=0 because it is at the peak of its arc, and t is unknown. The most important point here to understand is that the displacement s(1) of the first particle is not just s(1) = -4.9t^2. Because the particle starts with a displacement of 22.5, what you're trying to calculate is the difference between its original displacement and its final displacement. Therefore, the equation for the displacement, denoted by s(1), is given by the following:

s(1) = 22.5 - 4.9t^2

Looking at the second particle, u=15(as given in the question), and so the equation for the displacement, s(2), of the second particle is given by:

s(2) = 15t - 4.9t^2

Now, when the two particles collide, their displacements will be equal to one another. Therefore we can say the following:

s(1) = s(2)
22.5 - 4.9t^2 = 15t - 4.9t^2

Cancelling out the two 4.9t^2's gives the following:

22.5 = 15t
t = 1.5

And there you have it, you have proven that the point of collision between the two particles (ie. the time when they have equal displacement to O) is at time t=1.5, which is what the question was asking.

Edit: I forgot to mention how to do the second part of the question, finding the height above O where the collision takes place. This is simply done by plugging in the value of t=1.5 into one of the earlier equations. I'll use the equation for the second particle, giving s = 22.5 - 4.9*2.25 = 11.475m above O.