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# M2 Collisions watch

1. Q11.After the collision, how do I know which velocity is greater? If v1 and v2 are the respective velocities:
1) If v1>v2, I get v1 = 30, v2 = 21
S = 21 * 5 = 105
105 = v1 * t1
t1 = 3.5 s
2) If v2 > v1, I get v1 = 15, v2 = 24
S = 24 * 5 = 120
120 = v1 * t1
t1 = 6 s

The book gives 3 s as the answer. And in general, if I didn't know that the first particle will meet the second one after the collision, shouldn't I also consider the case when the first particle changes direction? What am I missing?
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2. (Original post by to4ka)
Q11.After the collision, how do I know which velocity is greater? If v1 and v2 are the respective velocities:
1) If v1>v2, I get v1 = 30, v2 = 21
S = 21 * 5 = 105
105 = v1 * t1
t1 = 3.5 s
2) If v2 > v1, I get v1 = 15, v2 = 24
S = 24 * 5 = 120
120 = v1 * t1
t1 = 6 s

The book gives 3 s as the answer. And in general, if I didn't know that the first particle will meet the second one after the collision, shouldn't I also consider the case when the first particle changes direction? What am I missing?
The text of the question says 2kg and 10 kg are travelling in the same direction initially. The 2kg mass catches up with the 10 kg mass, and they collide.

Does that cover your misunderstanding?

Edit: There is only one possible set of values for v1,v2. How come you have two? EndEdit.

You also made a siip at the end "120 = v1 * t1" implies v1 = ?? Not 6.
And note that they want the additional time, not just the time since the impact.
3. I just realized v1 > v2 makes no sense. And if the first particle changes direction, energy is created... but shouldn't the two cases be discussed when solving such problems?

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4. (Original post by ghostwalker)
The text of the question says 2kg and 10 kg are travelling in the same direction initially. The 2kg mass catches up with the 10 kg mass, and they collide.

Does that cover your misunderstanding?
Does "collide" imply they don't change directions?

You also made a siip at the end "120 = v1 * t1" implies v1 = ?? Not 6.
t1 is the time between B being brought to rest and A reaching B for the second time. B travels with 24 ms-1 for 5 s, which is 120 m, and then is brought to rest. By that moment, A has travelled 5 * 15 = 75 m, so there are (120-75)÷15 = 3 s for A to catch up. Thank you, got it!

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5. (Original post by to4ka)
Does "collide" imply they don't change directions?
No. "Collide" doesn't tell you anything about their movement after the collision. But your conservation of momentum, and restitution equations (once you've solved them) do tell you that both velocities are positive (assuming you defined them in the same direction as before) after collision, and hence are both still moving in the same direction.
6. And what about the case when A changes direction? I know it's impossible in this question as you're told they collide again, but if the question was shorter and asked for the final velocities only, shouldn't we consider the case?

Because, if A changes direction:
e = (v1 + v2) / 15
and since by cons. of momentum
270 = -v1 * 2 + 10 * v2
I get
v2 = 31,5
v1 = - 22.5

which is impossible as spd is positive. But why is this case ignored? Shouldn't I prove that the particles don't change direction of motion?
7. (Original post by to4ka)
I just realized v1 > v2 makes no sense. And if the first particle changes direction, energy is created... but shouldn't the two cases be discussed when solving such problems?

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There will only be one case. Having defined your directions for v1 and v2, you will only get one set of numbers, positive or negative depending on the situation and the directions you defined v1,v2 in.

You can tell from the wording of the question that v1,v2 are both going to be in the same direction as previous.

If you inadvertently defined v1 in the opposite direction, it's not a problem as the value you'd get would be negative showing that the direction is actually opposite to the one you assumed.
8. Your equations do not tell you particles don't change direction - you assume it by writing
e = (v2 - v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15

EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2 - v1 or v1 + v2.

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9. (Original post by to4ka)
Your equations do not tell you particles don't change direction - you assume it by writing
e = (v2 - v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15

EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2 - v1 or v1 + v2.

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Sorry, this is taking longer than I have time for right now. I'll have to leave it for someone else to pick up. If it's not sorted, I'll post again tomorrow.
10. (Original post by to4ka)
Your equations do not tell you particles don't change direction - you assume it by writing
e = (v2 - v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15

EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2 - v1 or v1 + v2.

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You can't "define v1 and v2 as speed" - you are using them in a momentum equation which is a vector equation (momentum is a vector quantity).

If you write v2 - v1, then your answer will be (let's say) v_1 = 20 and v_2 = 30.

had you written v_2 + v_1, your answer would be v_1 = -20 and v_2 = 30.

In both cases, you'd have gotten 30 - 20 = 10.

i.e: just use one set and the algebra will sort itself out and give you a positive/negative value depending on whether your assumed direction is the same as the actual direction. If your answer is positive, the assumed direction is the same as the actual direction. If your answer is negative, the assumed direction is the opposite of the actual direction.
11. (Original post by Zacken)
You can't "define v1 and v2 as speed" - you are using them in a momentum equation which is a vector equation (momentum is a vector quantity).

If you write v2 - v1, then your answer will be (let's say) v_1 = 20 and v_2 = 30.

had you written v_2 + v_1, your answer would be v_1 = -20 and v_2 = 30.

In both cases, you'd have gotten 30 - 20 = 10.

i.e: just use one set and the algebra will sort itself out and give you a positive/negative value depending on whether your assumed direction is the same as the actual direction. If your answer is positive, the assumed direction is the same as the actual direction. If your answer is negative, the assumed direction is the opposite of the actual direction.
Right, this is all good, but I get stuck in the middle:

Let (->) be positive, define v1 and v2 as the final velocities of A and B.

By conservation of momentum, we have:
35 * 2 + 20 * 10 = v1 * 2 + v2 * 10
so
v1 = 135 - 5v2
We also have that
e = (relative speed of separation) / (relative speed of approach) = ??? / 15

What is ??? - surely it can't be (v2 - v1) as this is relative velocity; I need the spd. It must be |v2-v1| = |6v2 - 135|

which gives
|6v2-135| = 9
so v2 is either 21 or 24. In the first case I get v1 = 30 > v1 which is impossible and the second case is all good. Is this right? Or is there a simpler way?

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12. (Original post by to4ka)
Right, this is all good, but I get stuck in the middle:

Let (-> be positive, define v1 and v2 as the final velocities of A and B.

By conservation of momentum, we have:
35 * 2 + 20 * 10 = v1 * 2 + v2 * 10
so
v1 = 135 - 5v2
We also have that
e = (relative speed of separation) / (relative speed of approach) = ??? / 15

What is ??? - surely it can't be (v2 - v1) as this is relative velocity; I need the spd. It must be |v2-v1| = |6v2 - 135|

which gives
|6v2-135| = 9
so v2 is either 21 or 24. In the first case I get v1 = 30 > v1 which is impossible and the second case is all good. Is this right? Or is there a simpler way?

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e = (relative velocity of separation) / (relative velocity of approach)

So, indeed, it is (v2 - v1)/15.
13. Uhm, not? Relative velocity is a vector and you can't divide vectors (not at this point, at least) by vectors.

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14. (Original post by to4ka)
Uhm, not? Relative velocity is a vector and you can't divide vectors (not at this point, at least) by vectors.

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Time is displacement/velocity and they're both vectors, aren't they?
15. Techincally,

time = |displacement|/|velocity|

as in, if velocity is 3i+4j and displacement is 15j

time isn't (15j)/(3i+4j), it's 15 / sqrt(3*3+4*4), which is 3.

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16. Posted from TSR Mobile
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17. (Original post by to4ka)
Techincally,

time = |displacement|/|velocity|

as in, if velocity is 3i+4j and displacement is 15j

time isn't (15j)/(3i+4j), it's 15 / sqrt(3*3+4*4), which is 3.

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Yeah, sorry - my bad!

It's just that relative speed = difference in velocities.
18. Are you sure?

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19. (Original post by to4ka)
...
Doesn't look as if things are resolved, so:

Two cases:

1. We take v1 and v2 going in the same direction as the initial velocities.

By momentum we have: and so:

By restitution:

"speed separation" = e( "speed of approach" )

Hence,

(1)+(2) gives

And thus , and subbing into (2) we get

2. We take v1 going back in the opposite diretion and v2 going in the same direction as the intial velocities.

By momentum we have: and so:

By restitution:

"speed separation" = e( "speed of approach" )

Hence,

(3)+(4) gives

And thus , and subbing into (4) we get
Since we assumed v1 was going back in the opposite direction, and we get a minus value, then v1 is in fact going in the original direction, same as all the other velocities.
20. Thanks, I understood quite a lot from you both!

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