# Problem 8-43 2 dim. collision - Part 2 - D

Two balls of equal mass m undergo a collision. One ball is initially stationary. After the collision, the velocities of the balls make angles of $31.1^\circ$ and $48.9^\circ$ relative to the original direction of motion of the moving ball.

(a) Draw a diagram showing the initial and final situations. If you are uncertain about the final directions of motion, remember that momentum is conserved.

(b) If the initial speed of the moving ball is $2.25\; m/s$, what are the speeds of the balls after the collision?

[Ans. (a) $1.18\; m/s$ at $48.9^\circ$;   (b) $1.72 \;m/s$ at $31.1^\circ$    (c) $no$]

Accumulated Solution

Correct!

From conservation of momentum

$2.25 = v{_1}' \cos 48.9 + v{_2}' \cos 31.1 \\ 2.25 = (0.6574) v{_1}' + (0.8563) v{_2}' \; \text {Eqn.# 1}$

The "before" and "after" $y$-components of momentum are:

Solution Before After
(A) $2.25\;m$ $m(v{_1}' \sin 48.9 - v{_2}' \sin 31.1)$
(B) $2.25\;m$ $m(v{_1}' \cos 48.9 - v{_2}' \cos 31.1)$
(C) $0$ $m(v{_1}' \sin 48.9 - v{_2}' \sin 31.1)$
(D) $0$ $m(v{_1}' \sin 48.9 + v{_2}' \sin 31.1)$