Question

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially traveling at 14.6 and is deflected 28.0 from its initial direction. Assume that the collision is perfectly elastic.
a)Find the final speed of the puck B after the collision.
b)Find the final speed of the puck A after the collision.
c)Find the direction of B's velocity after the collision.

Answer 1

Answer:

$a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree$

Explanation:

Puck A's initial speed is $v_a_1=14.6m/s$ and move in a direction $\theta_b=28.0\textdegree$ after the collision.

#$P_1=P_2$ since there's no external force on the system($P=mv).$

#The collision equation can be written as;

$m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2$

The kinetic energies before and after the collision are expressed as:

$K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2$

$0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2$

Let +x along A's initial direction and +y along A's final direction makes the angle $62\textdegree$

$\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j$

#Substitute in $v_{b2}^2=213.16-v_{a2}^2$:

$\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}$

$v_{a2}=12.891\ m/s$

Hence, the the speed of puck A after the collision is 12.891  m/s

#. The velocity of A after the collision is;

$\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j$

Substitute $\dot v_{a2}$ into $\dot v_{b2}=(14.6i)-\dot v{a2}$:

$\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j$

This is the velocity of puck B after the collision, it's speed is:

$v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s$

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

$\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree$

Hence, the direction of B's velocity after the collision is 62°