6. First conserve momentum (the system is closed and isolated). Let M=100kg and v = 3 m/s. Let u be the speed of the final m+M system. Then
M v = (m+M) u.
Since 20% of the initial kinetic energy is lost, conservation of total energy gives us
0.8 M/2 v2 = (m+M)/2 u2.
Solving the first equation for u, substituting into the second equation, and rearranging gives
m = M/0.8 - M = 25 kg.
7. Let u1 be the speed of particle 1 after the collision, similarly for particle 2.
Conserving momentum gives
m v = m u1 + M u2
where the velocities are vectors. Break this into its components:
m v = m u1 cos(-25) + M u2 cos(15)
0 = m u1 sin(-25) + M u2 sin(15)
m v2 = m u12 + M u22.
Solve the second conservation of energy equation to get
u1 = M u2 sin(15)/[ m sin(25)]
and the first to get
v = M/m [sin(15)/tan(25) + cos(15)] u2
Now solve the conservation of energy equation to get m:
m = M[ (sin(15)/tan(25)+cos(15))2 - sin(15)2/sin(25)2] = 552 gm.