Answer:
[tex]a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree[/tex]
Explanation:
Puck A's initial speed is [tex]v_a_1=14.6m/s[/tex] and move in a direction [tex]\theta_b=28.0\textdegree[/tex] after the collision.
#[tex]P_1=P_2[/tex] since there's no external force on the system([tex]P=mv).[/tex]
#The collision equation can be written as;
[tex]m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2[/tex]
The kinetic energies before and after the collision are expressed as:
[tex]K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2[/tex]
[tex]0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2[/tex]
Let +x along A's initial direction and +y along A's final direction makes the angle [tex]62\textdegree[/tex]
[tex]\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j[/tex]
#Substitute in [tex]v_{b2}^2=213.16-v_{a2}^2[/tex]:
[tex]\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\[/tex]
[tex]v_{a2}=12.891\ m/s[/tex]
Hence, the the speed of puck A after the collision is 12.891 m/s
#. The velocity of A after the collision is;
[tex]\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j[/tex]
Substitute [tex]\dot v_{a2}[/tex] into [tex]\dot v_{b2}=(14.6i)-\dot v{a2}[/tex]:
[tex]\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j[/tex]
This is the velocity of puck B after the collision, it's speed is:
[tex]v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s[/tex]
The velocity of puck B after the collision is 6.8544 m/s
c. The direction of puck B after the collision is:
[tex]\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree[/tex]
Hence, the direction of B's velocity after the collision is 62°
