Answer
Puck A has a final speed of $3.43m/s$ and puck B has a final speed of $2.6m/s$
Work Step by Step
We examine the collisions in the x and y components separately.
a) x Component:
Puck A, whose mass is $m=0.025kg$ and has initial horizontal velocity $\vec{v}_{0x}=+5.5m/s$, collides with puck B, whose mass is $M=0.05kg$ and has initial velocity $\vec{V}_{0x}=0$.
After the collision, puck A has a horizontal velocity $\vec{v_{fx}}=\vec{v_f}\cos65$ and puck B has a horizontal velocity $\vec{V_{fx}}=\vec{V_f}\cos37$
We assume there are no external forces, so the total linear momentum is conserved. Therefore, $$M\vec{V}_{0x}+m\vec{v}_{0x}=M\vec{V}_{fx}+m\vec{v}_{fx}$$ $$0.05\vec{V_f}\cos37+0.025\vec{v_f}\cos65=0+0.025(+5.5) $$ $$0.05\vec{V_f}\cos37+0.025\vec{v_f}\cos65=0.14$$ $$2\vec{V_f}\cos37+\vec{v_f}\cos65=5.6$$ $$1.6\vec{V_f}+0.42\vec{v_f}=5.6 (1)$$
b) y Component:
Both puck A and puck B have zero initial vertical velocity, so $\sum\vec{p_{0y}}=0$.
After the collision, puck A has vertical velocity $\vec{v_{fy}}=\vec{v_f}\sin65$ and puck B has vertical velocity $\vec{V_{fy}}=-\vec{V_f}\sin37$
We assume there are no external forces, so the total linear momentum is conserved. Therefore, $$M\vec{V}_{fy}+m\vec{v}_{fy}=0$$ $$-0.05\vec{V_f}\sin37+0.025\vec{v_f}\sin65=0$$ $$-2\vec{V_f}\sin37+\vec{v_f}\sin65=0$$ $$-1.2\vec{V_f}+0.91\vec{v_f}=0 (2)$$
Solve (1) and (2), we have $V_f=2.6m/s$ and $v_f=3.43m/s$
