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(a)guyz anyone can help with this its pretty easy just not to my brain r8 now .. here is it just part b ! =p
Two particles A and B are moving on a smooth horizontal plane. The mass of A is km,
where 2 <
k < 3 , and the mass of B is m. The particles are moving along the same straight
line, but in opposite directions, and they collide directly. Immediately before they collide
the speed of
A is 2u and the speed of B is 4u. As a result of the collision the speed of A is
halved and its direction of motion is reversed.
(a) Find, in terms of
k and u, the speed of B immediately after the collision.
(b) State whether the direction of motion of
B changes as a result of the collision,explaining your answer.
m(A)u(A) + m(B)u(B) = m(A)v(A) + m(B)v(B)
(km)(2u) + (m)(-4u) = (km)(-u) + (m)v(B)
Eliminating m from both sides
2ku - 4u = -ku + v(B)
3ku - 4u = v(B)
v(B) = u(3k - 4)
(b)
We know that k lies in between 2 and 3, so lets assume k = 2.5
v(B) = u(3(2.5) - 4)
v(B) = u(7.5 - 4)
v(B) = 3.5u
v(B) is positive
and it was initially negative so
the direction changes