Recent content by arizona1379

However we will not provide it. lol

:cry:

Break this down into your knowns and unknowns. You only have four kinematic equations. This problem is a projectile problem that is asking you to break things up in terms of x and y. First the shell moves initially with the bullet at 300 m/s horizontally, and then when it approaches 0...

Finalllyyyyyyyy:!)

AHA.:smile: ΔKEcue= ((.5)(0.26 kg)(1.2m/s)2+(.5)(0.15kg)(0)2)i-((.5)(0.26 kg)(-0.32 m/s)2)f =0.1872J-0.0133J =.1739J 100(.1739J)/.1872= 93%

How do I know that they are not? Lol I guess I just assumed that it's not a glancing collision. I thought the two balls bounce off each other in opposite directions, so the vf1 would be negative. No angles were given.

(0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2) (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2) ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2) (2.08 m/s)+1.73(vf1)=(vf2) (v1-v2)i=(v2-v1)f 1.2-(vf1)=vf2...

How could both of my final velocities be positive if they are traveling in two different directions after collision?

Yeah I guess I need practice. I am having a hard time finding a good example. I haven't had math in a while so my algebra isn't very strong.

(0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2) (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2) ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2) (2.08 m/s)+1.73(vf1)=(vf2) (1.2 m/s)+vf1=vf2 (2.08...

Providing the equation would have been less effort on both our parts. Thank you for the help. :)

((.5)(0.26kg)(1.2 m/s)2+(.5)(0.15kg)(0)2)i = ((.5)(0.26kg)(vf1)2+(.5)(0.15kg)(Vf2)2)f (0.1872 kg m/s)+0=(0.13kg)(-vf1)2+(0.075kg)(vf2) (0.1872 kg m/s)+(0.13kg)(vf1)2=(0.075kg)(vf2)2 SQR[(0.1872 kg m/s)+(0.13kg)(vf1)2]/(0.075kg)=vf (2.895 m/s)(vf1)=VF2...

(0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(vf2) (0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(vf2) (0.312 kg m/s)+(0.26kg)(vf1)= (0.15kg)(vf2) ((0.312 kg m/s)+(0.26kg)(vf1))/(0.15kg)=(vf2) ((.5)(0.26kg)(1.2m/s)2+(.5)(0.15kg)(0 m/s)2)i =...

Oh so I need to use ΔKE equation. ((.5)(m1)(Vi1)2+(.5)(m2)(Vi2)2)i = ((.5)(m1)(Vf1)2+(.5)(m2)(Vf2)2)f and solve for either Vf1 or Vf2 then plug it back into: (m1v2+m2v2)i=(m1v2+m2v2)f Find the other final velocity, and then: ΔKE= ((.5)(m1)(Vi1)2+(.5)(m2)(Vi2)2)i...

Homework Statement A 0.26 kg cue ball with a velocity 1.2 m/s collides elastically with a 0.15 kg billiard ball at rest. What percentage of the initial kinetic energy is transferred to the billiard? m1= 0.26kg Vi1= 1.2 m/s Vf1= ? m2= 0.15kg Vi2= 0 m/s Vf2= ? Homework Equations...