A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
A billiard ball moving at 2.2 m/s strikes a second billard ball at rest and continues at 45 degrees from its original direction, while the other moves at 45 degrees from the same original direction. Both billiard balls have the same mass. Find the speed of the second billiard ball.
1.6 m/s, 3.1 m/s, 0.78 m/s, or 1.1 ,/s
**not sure! thank you!
anonymous
 one year ago
A billiard ball moving at 2.2 m/s strikes a second billard ball at rest and continues at 45 degrees from its original direction, while the other moves at 45 degrees from the same original direction. Both billiard balls have the same mass. Find the speed of the second billiard ball. 1.6 m/s, 3.1 m/s, 0.78 m/s, or 1.1 ,/s **not sure! thank you!

This Question is Closed

perl
 one year ago
Best ResponseYou've already chosen the best response.4can we make a diagram of the situation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, i am not quite sure how!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes:) what do we do from here?

perl
 one year ago
Best ResponseYou've already chosen the best response.4note that the collision is not 'head on' , thats why it goes off an angle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes:) so in calculating, would it be moving at the speed of 3.1 m/s?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! what do i plug in for m? :/

perl
 one year ago
Best ResponseYou've already chosen the best response.4momentum = mass *velocity In elastic collisions, momentum is conserved, therefore "initial momentum = final momentum" . Also kinetic energy is conserved. initial x momentum = 2.2*m + 0*m initial y momentum = 0*m + 0*m final x momentum = m(vf1 cos 45) + m*vf2 cos(45) final y momentum = m (vf1 sin 45)+ m (vf2 sin(45)) 2.2 m = m*vf1 cos 45 + m*vf2 cos(45) 0 = m (vf1 sin 45)+ m (vf2 sin(45)) 1/2m*2.2^2 = 1/2 m* vf1^2 + 1/2* m*vf2^2

perl
 one year ago
Best ResponseYou've already chosen the best response.4we get a system of three equations, (we can cancel m out)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! and so we get left with vf1 and vf2? we are looking for vf2 becuase of the second billiard\?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how can i solve that value?

perl
 one year ago
Best ResponseYou've already chosen the best response.4momentum = mass *velocity In elastic collisions, momentum is conserved, therefore "initial momentum = final momentum" . Also kinetic energy is conserved. initial x momentum = 2.2*m initial y momentum = 0 final x momentum = m(vf1 cos 45) + m*vf2 cos(45) final y momentum = m (vf1 sin 45)+ m (vf2 sin(45)) 2.2 m = m*vf1 cos 45 + m*vf2 cos(45) 0 = m (vf1 sin 45)+ m (vf2 sin(45)) 1/2m*2.2^2 = 1/2 m* vf1^2 + 1/2* m*vf2^2 canceling out mass m 2.2 = vf1 cos 45 +vf2 cos(45) 0 = (vf1 sin 45)+ (vf2 sin(45)) 1/2*2.2^2 = 1/2 vf1^2 + 1/2*vf2^2 from the second equation vf1 sin45 = vf2 sin(45) vf1 = vf2

perl
 one year ago
Best ResponseYou've already chosen the best response.4now plug that into the first equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay ...ermm i got 0.7071 :/ that doesn't look right though L:/

perl
 one year ago
Best ResponseYou've already chosen the best response.42.2 = vf1 cos 45 +vf2 cos(45) vf1 = vf2 2.2 = vf1 cos 45 +vf1 cos(45) because cos(45) = cos(45) 2.2 = 2vf1 * cos(45) 2.2 /( 2 cos45) = vf1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay 1.555 so our solution is 1.6 m/s ?

perl
 one year ago
Best ResponseYou've already chosen the best response.4to the tenth place , or 2 sig figs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay! thanks so much!

perl
 one year ago
Best ResponseYou've already chosen the best response.4so it turns out that the speed is the same for both balls after the collision. this is because they are both the same mass and diverge at symmetric angles

perl
 one year ago
Best ResponseYou've already chosen the best response.4nice, so it turns out we didn't have to use kinetic energy in this problem. but if they are different masses, then you have to use kinetic energy

perl
 one year ago
Best ResponseYou've already chosen the best response.4but it doesnt hurt to pull out all the equations , just in case

perl
 one year ago
Best ResponseYou've already chosen the best response.4any questions on the trig part?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, it all makes sense now :) thank you!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.