B BrianK0220 Full Member 10+ Year Member Joined Sep 24, 2010 Messages 123 Reaction score 0 Points 0 Jul 7, 2012 #1 Advertisement - Members don't see this ad Can anyone explain this? If it's elastic, then we use: m1v1=(m1+m2)v2 Arbitrary values: m1=1 v1=3 m2=3 v2=0.75 Wouldn't the answer be (1/4)v1? Thanks guys.
Advertisement - Members don't see this ad Can anyone explain this? If it's elastic, then we use: m1v1=(m1+m2)v2 Arbitrary values: m1=1 v1=3 m2=3 v2=0.75 Wouldn't the answer be (1/4)v1? Thanks guys.
Charles Darwin KFBR392 10+ Year Member Joined Jun 19, 2011 Messages 1,159 Reaction score 6 Points 4,591 Location CA, USA Non-Student Jul 7, 2012 #2 BrianK0220 said: Can anyone explain this? If it's elastic, then we use: m1v1=(m1+m2)v2 Arbitrary values: m1=1 v1=3 m2=3 v2=0.75 Wouldn't the answer be (1/4)v1? Thanks guys. Click to expand... If it's elastic, you have conservation of linear momentum and conservation of kinetic energy. You wrote the equation for a perfectly inelastic collision... EDIT: And yes, 1/4 of 3 is 3/4. Last edited: Jul 7, 2012 Upvote 0 Downvote
BrianK0220 said: Can anyone explain this? If it's elastic, then we use: m1v1=(m1+m2)v2 Arbitrary values: m1=1 v1=3 m2=3 v2=0.75 Wouldn't the answer be (1/4)v1? Thanks guys. Click to expand... If it's elastic, you have conservation of linear momentum and conservation of kinetic energy. You wrote the equation for a perfectly inelastic collision... EDIT: And yes, 1/4 of 3 is 3/4.
B BrianK0220 Full Member 10+ Year Member Joined Sep 24, 2010 Messages 123 Reaction score 0 Points 0 Jul 7, 2012 #3 That explains a lot, including why my answer was an answer choice. Thanks. Upvote 0 Downvote