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## Homework Statement

A steel ball of mass 0.50 kg, moving with a velocity of 2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially at rest.The collision is a glancing one, causing the moving ball to have a velocity of 1.5 m/s [E 30 N] after the collision. Determine the velocity of the second ball after the collision.

## Homework Equations

Momentum = final momentum

m1v1 +m2v2 = m1v1 +m2v2

(c)^2 = (a)^2 + (b)^2 - 2(a)(b)cos C

## The Attempt at a Solution

m1 = 0.50 kg

v1 = 2.0 m/s[E]

m2 = 0.30 kg

v2 = 0 m/s

v1 (final) = 1.5 m/s [E 30 N]

v2 (final) = ?

Ptotal = Ptotal (final)

m1v1 + m2v2 = m1v1 + m2v2

(0.50)(2.0[E]) = (0.50)(1.5[E 30 N]) + (0.30) (v2(final))

1.0 [E] = 0.75[E 30 N] + 0.30(v2)

**Diagram i drew**

http://img259.imageshack.us/my.php?image=diagramhe8.png

(c)^2 = (a)^2 + (b)^2 - 2(a)(b)cos C

((0.30)(V2))^2 = 0.75^2 + 1^2 -2(0.75)(1) cos 30

0.09V2^2 = 2.5625 - 0.866025403

V2 = square root(7.738606633)

i used sign law to find out the equation

V2(final) = 2.78[E 10.4 S]

After doing all this work i went to the back of the book and the answer was 1.7 m/s[E 47 S] i hate when this happens:(

if somebody can help me out i will really appreciate it

thanks alot