a.
The momentum is given by:
[tex]\begin{gathered} p=mv \\ where: \\ m=0.3kg \\ v=23m/s \\ so: \\ p=0.3(23) \\ p=6.9kg\frac{m}{s} \end{gathered}[/tex]b.
We can use the impulse formula:
[tex]\begin{gathered} F\cdot\Delta t=m\cdot\Delta v \\ so: \\ F=m\cdot\frac{\Delta v}{\Delta t} \\ \\ F=0.3\cdot\frac{23}{0.1} \\ F=69N \end{gathered}[/tex]c.
[tex]\begin{gathered} I=m\Delta v \\ where: \\ \Delta v=vf-vo=21-23=-2m/s \\ so: \\ I=0.3(-2) \\ I=-0.6kg\cdot m/s \end{gathered}[/tex]d.
Using conservation of momentum:
[tex]m1u1+m2v1=m1u2+m2v2[/tex]Where:
[tex]\begin{gathered} pi=m1u1+m2v1 \\ m1=0.3kg \\ u1=21m/s \\ m2=2kg \\ v1=0m/s \\ so: \\ pi=0.3(21)+2(0) \\ pi=6.3kg\cdot m/s \end{gathered}[/tex]e.
Now, we can solve for the right hand side of the equation of the conservation of momentum:
[tex]\begin{gathered} 6.3=m1u2+m2v2 \\ v2=\frac{6.3-m1u2}{m2} \\ where: \\ m1=0.3kg \\ m2=2kg \\ u2=4m/s \\ so: \\ v2=\frac{6.3-(0.3)(4)}{2} \\ v2=2.55m/s \end{gathered}[/tex]