# mechanics- collisionsWatch

#1
hi, i did this question on collision but i don't think i've got the right answer.

the question is:
x and y lie on a straight line 30m apart.
at time t=0 they are moving towards each other.
x has initial speed 15ms-1 and accelerates at rate of 1ms-2
y has initial speed 20ms-1 and accelerates at rate of 2ms-2

calculate the time it takes for them to collide

so i did 20t + t2 = 15t + 1/2t2 and rearranged
and ended up with t(t+10) = 0 which means t would have to equal -10... have i gone wrong somewhere?
0
1 week ago
#2
(Original post by entertainmyfaith)
hi, i did this question on collision but i don't think i've got the right answer.

the question is:
x and y lie on a straight line 30m apart.
at time t=0 they are moving towards each other.
x has initial speed 15ms-1 and accelerates at rate of 1ms-2
y has initial speed 20ms-1 and accelerates at rate of 2ms-2

calculate the time it takes for them to collide

so i did 20t + t2 = 15t + 1/2t2 and rearranged
and ended up with t(t+10) = 0 which means t would have to equal -10... have i gone wrong somewhere?
It looks to me as if you have worked out the distance that each has travelled in the time t, and put them equal to each other. But would they both travel the same distance? Not necessarily (and in fact, certainly not in this case).

If one of them travels a distance s, the other will have to travel a distance of 30 - s.
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#3
(Original post by Pangol)
It looks to me as if you have worked out the distance that each has travelled in the time t, and put them equal to each other. But would they both travel the same distance? Not necessarily (and in fact, certainly not in this case).

If one of them travels a distance s, the other will have to travel a distance of 30 - s.
ah, i see (thinking they were equal was a pretty poor assumption of mine). so would simultaneous equations not be appropriate in this case?
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1 week ago
#4
(Original post by entertainmyfaith)
ah, i see (thinking they were equal was a pretty poor assumption of mine). so would simultaneous equations not be appropriate in this case?
You do have to use simultaneous equations, it's just that you can't put the "s" for each particle equal to each other. You need to put "s" for one particle equal to "30 - s" of the other.
0
1 week ago
#5
(Original post by entertainmyfaith)
ah, i see (thinking they were equal was a pretty poor assumption of mine). so would simultaneous equations not be appropriate in this case?
Set x to be at some origin and y to be 30m away. s(x) is always 30 - s(y), so there's one unknown gotten rid of.
You should be able to cancel out some of your variables. One velocity will have to be negative because they're moving in opposite directions and they'll both have been travelling for the same time at collision.

Can't help much more, I'm on the bus atm
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#6
(Original post by Pangol)
You do have to use simultaneous equations, it's just that you can't put the "s" for each particle equal to each other. You need to put "s" for one particle equal to "30 - s" of the other.
i ended up with -24.1611 and 0.82777... these don't seem right
(Original post by Sinnoh)
Set x to be at some origin and y to be 30m away. s(x) is always 30 - s(y), so there's one unknown gotten rid of.
You should be able to cancel out some of your variables. One velocity will have to be negative because they're moving in opposite directions and they'll both have been travelling for the same time at collision.

Can't help much more, I'm on the bus atm
i'm sorry but you've confused me even more
0
1 week ago
#7
(Original post by entertainmyfaith)
i'm sorry but you've confused me even more
Yeah I realise now my first sentence was wrong af.
Have you sketched a diagram?
If the velocity of x is positive 15, you'll have to treat the velocity of y as negative; if they're moving towards each other, then they're moving in exactly opposite directions.

At the collision, their displacement from the "start" is equal. So if x has moved s metres, y has moved (30 - s) metres. In addition, the time for which they have travelled is also equal.
0
1 week ago
#8
(Original post by entertainmyfaith)
i ended up with -24.1611 and 0.82777... these don't seem right

i'm sorry but you've confused me even more
If you look at the velocities they won't take long to collide - can you double check the values?

I found s for both then added and put equal to 30.
0
1 week ago
#9
So: consider the motion separately for each particle. Find the function of the distance of y keeping the time in terms of t and do the same for x.

When they collide, the distances they have travelled will sum to 30 - because originally they were 30 metres apart, now they're 0 metres apart. So the displacement of y can be rewritten as 30 - the displacement of x.
That eliminates an unknown, leaving you with a quadratic in terms of t.
0
#10
(Original post by Muttley79)
If you look at the velocities they won't take long to collide - can you double check the values?

I found s for both then added and put equal to 30.
i keep plugging 3t2 +70t -60 = 0 into the quadratic formula and getting 0.827776542 (didn't state all the dp before whoops) so 0.828 to 3dp?
(Original post by Sinnoh)
So: consider the motion separately for each particle. Find the function of the distance of y keeping the time in terms of t and do the same for x.

When they collide, the distances they have travelled will sum to 30 - because originally they were 30 metres apart, now they're 0 metres apart. So the displacement of y can be rewritten as 30 - the displacement of x.
That eliminates an unknown, leaving you with a quadratic in terms of t.
i ended up adding them and equating it to 30
drawing the diagram really helped me though!!
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1 week ago
#11
(Original post by entertainmyfaith)
i ended up adding them and equating it to 30
drawing the diagram really helped me though!!
Same thing what answer did you get in the end?

Also, rule of thumb: in mechanics, if you've not been given one, draw a diagram
Last edited by Sinnoh; 1 week ago
0
#12
i got -24.something and 0.827776542

(Original post by Sinnoh)
Same thing what answer did you get in the end?

Also, rule of thumb: in mechanics, if you've not been given one, draw a diagram
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1 week ago
#13
(Original post by entertainmyfaith)
i got -24.something and 0.827776542
I can confirm both are correct. Yes, even the crazy negative one. Maths NEVER lies — even when you think it does. For the -24 thing, it will work if you... screw around with time a bit. Namely, make it run backwards.
0
1 week ago
#14
(Original post by sqrt -1)
[material deleted]
Perhaps you are unaware of the posting guideline that asks users not to post full solutions?
0
1 week ago
#15
Is that a thing? Oh well, away it goes, I guess.
0
1 week ago
#16
(Original post by entertainmyfaith)
i got -24.something and 0.827776542
Oh yah nvm that's correct, I made a mistake factorising
0
#17
(Original post by Sinnoh)
Oh yah nvm that's correct, I made a mistake factorising
thanks sinnoh
(Original post by sqrt -1)
I can confirm both are correct. Yes, even the crazy negative one. Maths NEVER lies — even when you think it does. For the -24 thing, it will work if you... screw around with time a bit. Namely, make it run backwards.
0
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