Answer:
The speed of the ball B is 6.4 m/s. The direction is 50 degrees counterclockwise.
Explanation:
Assuming the collision is elastic, use the conservation of momentum to solve this problem. The conservation law implies that:
[tex]m\vec v_{A0}+m\vec v_{B0} = m\vec v_{A1}+m\vec v_{B1}[/tex]
(the total momentum of the two balls is the same before (index 0) and after (index 1) the collision). Since B is stationary and A and B have the same mass, this simplifies to:
[tex]\vec v_{A0} = \vec v_{A1}+\vec v_{B1}[/tex]
and allows us to determine the velocity of ball B after the collision:
[tex]\vec v_{B1} = \vec v_{A0}-\vec v_{A1}[/tex]
The above involves vectors. Your problem suggests to use the component method, which I am assuming means solving the above equation separately along the x and y axes. Define x to align with the original line of motion of the ball A before the collision, and y to be perpendicular to x, pointing up:
[tex]v_{B1x} = v_{A0x}-v_{A1x}\\v_{B1y} = v_{A0y}-v_{A1y}[/tex]
We just need to compute the x- and y-components of the known velocity of the ball A. Drs. Sine and Cosine come to help here.
[tex]v_{A1x} = |v_{A1}|\cos 40^\circ\\v_{A1y} = |v_{A1}|\sin 40^\circ[/tex]
so
[tex]v_{B1x} = |v_{A0}|\cos 0^\circ-|v_{A1}|\cos 40^\circ=(10.0-7.7\cdot 0.77) \frac{m}{s}\approx 4.1\frac{m}{s}\\v_{B1y} = |v_{A0}|\sin 0^\circ-|v_{A1}|\sin 40^\circ=(0-7.7\cdot 0.64) \frac{m}{s}\approx -4.9\frac{m}{s}[/tex]
The speed of the ball B is [tex]|v_{B1}| = \sqrt{4.1^2+(-4.9)^2}\frac{m}{s}\approx 6.4 \frac{m}{s}[/tex]. The direction (angle from horizontal) is [tex]\beta = \arcsin (-\frac{4.9}{6.4})\approx -50^\circ[/tex], i.e., 50 degrees counterclockwise.
