# Momentum

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#1
All I’m doing today is getting stuck on maths qs😕.

I’ve set up so many different equations but I just donno how to get to the answer
0
#2
that’s what I’ve done so far without all the extra work
0
1 year ago
#3
...
m1 > m2: um1 - um2 = m1v1 + m2v2 = 2m1v1
0.5 = (v2 - v1)/2u => v2 - v1 = u => v2 = v1 + u

v1 in terms of m1, m2: m1v1 = m2(v1 + u)
=> v1(m1 - m2) = um2 => v1 = um2/(m1 - m2)

So u(m1 - m2) = 2u.m1m2/(m1 - m2)
=> (m1 - m2)^2 = 2m1m2
=> m1^2 - (4m2)m1 + m2^2 = 0

Quad eqn or complete square for m1 in terms of m2, take +ve sqrt for m1 > m2, gives m1 = m2(2 + sqrt3).
Last edited by Physics Enemy; 1 year ago
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#4
(Original post by Physics Enemy)
m1 > m2: um1 - um2 = m1v1 + m2v2 = 2m1v1
0.5 = (v2 - v1)/2u => v2 - v1 = u => v2 = v1 + u

v1 in terms of m1, m2: m1v1 = m2(v1 + u)
=> v1(m1 - m2) = um2 => v1 = um2/(m1 - m2)

So u(m1 - m2) = 2u.m1m2/(m1 - m2)
=> (m1 - m2)^2 = 2m1m2
=> m1^2 - (4m2)m1 + m2^2 = 0

Quad eqn or complete square for m1 in terms of m2, take +ve sqrt for m1 > m2, gives m1 = m2(2 + sqrt3).
Really appreciate the help but how do u solve the quadratic with m1 and m2 involved?
0
1 year ago
#5
Really appreciate the help but how do u solve the quadratic with m1 and m2 involved?
Notation is probably throwing you off. Think of solving x^2 - 4kx + k^2 = 0 (m1 = x, m2 = k) i.e) x in terms of k.
Last edited by Physics Enemy; 1 year ago
0
#6
(Original post by Physics Enemy)
Notation is probably throwing you off. Think of solving x^2 - 4kx + k^2 = 0 (m1 = x, m2 = k) i.e) x in terms of k.
Yh it was but I’ve got it now. Thanks a lot 😁
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