Posted by celeste on Tuesday, November 27, 2012 at 1:34pm.
Such collisions are not elastic.
Are you sure they are not asking about an inelastic collision?
thats the exact question. I am trying to help my daughter, her answer does not seem right to me she got 100 kg. Do you know how to solve this problem step by step. ty
36.4
I will give you an outline of the derivation:
call v0 the speed of the first ball before collision, v1 and v2 the speeds of the first and second balls after collision
m1, m2 are the masses of the first and second balls
start with momentum conservation:
m1 v0 = m1 v1 + m2 v2
rewrite as m2 v2 = m1(v0-v1)
energy conservation gives you
1/2 m 1 v0^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2
divide through by 1/2 and rewrite:
m2 v2^2 = m1(v0^2-v1^2) = m1 (v0-v1)(v0+v1)
but from momentum we know that m2 v2 = m1(v0-v1), so substitute this into the energy equation and get
m2 v2^2 = m2 v2 (v0+v1) or v2 = v0+v1
substitute this back into the momentum equation and get that
v1 = (m1-m2)v0/(m1+m2)
which then yields
v2 = 2 m1 v0/(m1+m2)
in this case, v0=8m/s, m2 = 80kg, and v2=5m/s and we want to find m1
we have (using the v2 equation):
5m/s = 2 m1 x 8m/s/(80+m1)
solving for m1:
400 + 5m1 = 16m1
400 = 11m1
m1 = 36.4kg