2-d momentum question

  • Thread starter dmitrip
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  • #1
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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
 

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
19,261
2,757
Separate it into N-S (y) and E-W (x) components. The net momentum in the y direction = 0, since initially there is no momentum in the y or transverse direction.

Also consider conservation of energy.
 
  • #3
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can somebody please show me how extacly i can solve this problem using components, i dont know where to start:frown:
 

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