prove that bodies of identical masses exchange their velocities after head on elastic collision
Answers
Answered by
81
in elastic collision,
v2 - v1 = u1 - u2
so....... v2 = u1 - u2 + v1......(1)
from coservation of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2
but since masses are equal,
mu1 + mu2 = mv1 + mv2
substituting the value for v2 from (1)
m(u1 + u2) = m(v1 + u1 - u2 +v1)
mu1 + mu2 = 2mv1 + mu1 - mu2
2mu2 = 2mv1
u2 = v1
substituting this value of v1 in (1) we get,
v2 = u1 - u2 + u2
v2 = u1
as u can see, the initial velocity of one body has now become the final velocity of second body after collision and
initial velocity of the second body has becum the final velocity of the first.
v2 - v1 = u1 - u2
so....... v2 = u1 - u2 + v1......(1)
from coservation of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2
but since masses are equal,
mu1 + mu2 = mv1 + mv2
substituting the value for v2 from (1)
m(u1 + u2) = m(v1 + u1 - u2 +v1)
mu1 + mu2 = 2mv1 + mu1 - mu2
2mu2 = 2mv1
u2 = v1
substituting this value of v1 in (1) we get,
v2 = u1 - u2 + u2
v2 = u1
as u can see, the initial velocity of one body has now become the final velocity of second body after collision and
initial velocity of the second body has becum the final velocity of the first.
Answered by
17
Answer:
Explanation:
in elastic collision,
v2 - v1 = u1 - u2
so....... v2 = u1 - u2 + v1......(1)
from coservation of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2
but since masses are equal,
mu1 + mu2 = mv1 + mv2
substituting the value for v2 from (1)
m(u1 + u2) = m(v1 + u1 - u2 +v1)
mu1 + mu2 = 2mv1 + mu1 - mu2
2mu2 = 2mv1
u2 = v1
substituting this value of v1 in (1) we get,
v2 = u1 - u2 + u2
v2 = u1
as u can see, the initial velocity of one body has now become the final velocity of second body after collision and
initial velocity of the second body has becum the final velocity of the first.
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