A 3.7kg block moving at 2.0 m/s toward the west on a frictionless surface has an elastic head-on collision with a second 0.80kg block traveling east at 3.0 m/s.
3 answers
Determine the final velocity of first block. Assume due east direction is positive.
and?
momentum:
3.7*2+.80*3=3.7V1 + .80V2
energy:
1/2 2.7*2^2+1/2 .80*3^2=1/2 2.7V1^2+1/2 .8*V2^2
ok, solve for V2 in the first equation, in terms of V2. Then square it...a bit of paper involved.
Now insert that value of V2^2 into the second equation, and solve for V1. Note is a quadratic, so you will have to use the quadratic equation.
All this is a paper and pencil task, so have plenty of paper.
3.7*2+.80*3=3.7V1 + .80V2
energy:
1/2 2.7*2^2+1/2 .80*3^2=1/2 2.7V1^2+1/2 .8*V2^2
ok, solve for V2 in the first equation, in terms of V2. Then square it...a bit of paper involved.
Now insert that value of V2^2 into the second equation, and solve for V1. Note is a quadratic, so you will have to use the quadratic equation.
All this is a paper and pencil task, so have plenty of paper.
1 answer