# M2 Collisions HelpWatch

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#1
Hey, I really don't understand what to do next.

On Jan 2003 M2 Q6b
I've got as far as doing; 10/9u - 8/9ue = W (taking W as the speed after the second collision)
And i know what the speed is when B comes off the wall
How do i find e
0
14 years ago
#2
P and Q heading directly towards each other.
Cons linear momentum:
(2m)(5u/9) - (m)(8eu/9) = (2m)(0) + mV
10u/9 - 8eu/9 = V (1)

Using the law of restitution:
(1/3) = V/(5u/9 + 8eu/9)
5u/9 + 8eu/9 = 3V (2)
Multiplying (1) by 3 gives:
10u/3 - 8eu/3 = 3V (3)

Equating 2 and 3 gives:
5u/9 + 8eu/9 = 10u/3 - 8eu/3
32eu/9 = 25u/9.
32e = 25
e = 25/32.
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#3
(Original post by Gaz031)
P and Q heading directly towards each other.
Cons linear momentum:
(2m)(5u/9) - (m)(8eu/9) = (2m)(0) + mV
10u/9 - 8eu/9 = V (1)

Using the law of restitution:
(1/3) = V/(5u/9 + 8eu/9)
5u/9 + 8eu/9 = 3V (2)
Multiplying (1) by 3 gives:
10u/3 - 8eu/3 = 3V (3)

Equating 2 and 3 gives:
5u/9 + 8eu/9 = 10u/3 - 8eu/3
16eu/9 = 25u/9.
16e = 25
e = 25/16.

the ms has 25/32

and the closest i've got to it is;
5/9u + 8/9ue
-------------
10/9u + 8/9ue

which gives me 0.72 and the ms says 0.7812
0
14 years ago
#4
Pardon the error. I misread 8eu/9 as 8eu/3, i think.
I usually do things on paper first. I dislike working through maths with keyboard text.
0
#5
(Original post by Gaz031)
Pardon the error. I misread 8eu/9 as 8eu/3, i think.
I usually do things on paper first. I dislike working through maths with keyboard text.
im taking the value of V as minus....is that okay? (when it comes off the wall)
0
14 years ago
#6
(Original post by ThugzMansion7)
im taking the value of V as minus....is that okay? (when it comes off the wall)
Yes, as it's going in the opposite direction to originally.
0
#7
(Original post by Gaz031)
Yes, as it's going in the opposite direction to originally.
okay im getting a bit mixed up with my signs...i'll post what i've done in a sec....it's prob just a sign or two...
0
#8
(Original post by ThugzMansion7)
okay im getting a bit mixed up with my signs...i'll post what i've done in a sec....it's prob just a sign or two...

1 / 3 = -10/9u - 8/3ue / 5/9u - 8/9ue
0
#9
(Original post by ThugzMansion7)
1 / 3 = -10/9u - 8/3ue / 5/9u - 8/9ue
0
14 years ago
#10
Gaz031 did answer u, see his lastest edit above
0
#11
10u/3 - 8eu/3

shudnt that be 10u/3 + 8eu/3? why is it, what it is above?
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#12
(Original post by ThugzMansion7)
10u/3 - 8eu/3

shudnt that be 10u/3 + 8eu/3? why is it, what it is above?
any1?
0
14 years ago
#13
(Original post by ThugzMansion7)
1 / 3 = -10/9u - 8/3ue / 5/9u - 8/9ue
Is this for the 2nd collision?

At the 2nd collision, the vel of P is zero.
0
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