Asked by Anonymous
A 5.00 g object moving to the right at +20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at rest.
(a) Find the velocity of each object after the collision.
5.00 g object cm/s
10.0 g object cm/s
(b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object.
%
(a) Find the velocity of each object after the collision.
5.00 g object cm/s
10.0 g object cm/s
(b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object.
%
Answers
Answered by
Damon
momentum before = 5*20 = 100
momentum after = 100 = 5 v1 + 10 v2
Ke before = (1/2)5(400) = 1000
Ke after = 1000 = .5*5*v1^2 + .5*10*v2^2
or
2000 = 5 v1^2 + 10 v2^2
but v2 = (100-5v1)/10 = (10 -.5v1)
v2^2 = 100-10v1+.25v1^2
so
2000 = 5v1^1 + 10(100-10v1+.25v1^2)
solve quadratic for v1. That is it, go back and get v2 and the kinetic energies.
Now we did that all in cgs system rather than meters, kilograms, seconds. I assume you are using an old text. Your energies will be in dynes rather than joules for that reason.
momentum after = 100 = 5 v1 + 10 v2
Ke before = (1/2)5(400) = 1000
Ke after = 1000 = .5*5*v1^2 + .5*10*v2^2
or
2000 = 5 v1^2 + 10 v2^2
but v2 = (100-5v1)/10 = (10 -.5v1)
v2^2 = 100-10v1+.25v1^2
so
2000 = 5v1^1 + 10(100-10v1+.25v1^2)
solve quadratic for v1. That is it, go back and get v2 and the kinetic energies.
Now we did that all in cgs system rather than meters, kilograms, seconds. I assume you are using an old text. Your energies will be in dynes rather than joules for that reason.
Answered by
Lucy
thankyou it was really helpful
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