1. (a)

2. (d)

3. (b)

4. (a)

5. (d)

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 v^{2} = (m+M)/2 u^{2}.

Solving the first equation for u, substituting into the second equation, and rearranging gives

m = M/0.8 - M = 25 kg.

7. Let u_{1} be the speed of particle
1 after the collision, similarly for particle 2.

Conserving momentum gives
m v = m u_{1} + M u_{2}

where the velocities are *vectors*. Break this into its components:

m v = m u_{1} cos(-25) + M u_{2} cos(15)

0 = m u_{1} sin(-25) + M u_{2} sin(15)

Conserve energy:

m v^{2} = m u_{1}^{2} + M u_{2}^{2}.

Solve the second conservation of energy equation to get

u_{1} = M u_{2} sin(15)/[ m sin(25)]

and the first to get

v = M/m [sin(15)/tan(25) + cos(15)] u_{2}

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.