AL APPLIED

[匿名]

2011-08-14 3:48 am

A wedge, of mass M and angle alpha ,is initially at rest on a smooth horizontal table.A particle of mass m,initially at the bottom of the wedge, is projected with speed u up the face of the wedge along the line of greatest slope.The contact between the paticle and the wedge is also smooth .Assuming that the particle remains on the face of the wedge throughout this motion,determine the displacement of the wedge when the particle again reaches the bottom of the wedge.

回答 (1)

用戶25975

2011-08-15 5:55 pm

✔ 最佳答案

With ref. to the diagram below when the particle is on the wedge:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Aug11/Crazyapp2.jpg

Consider the vertical component of the ball's motion:

Acc. = g(1 - cos2 α) = g sin2 α downward.

Initial vertical velocity = u sin α upward

So applying the equation in vertical direction:

0 = (u sin α)t - (g sin2 α)t2/2 since when the ball returns to the bottom, vertical displacement = 0.

(g sin2 α)t/2 = u sin α

t = 2u/(g sin α)

which is the time elapsed before the ball returns to the bottom.

Then consider the wedge's motion:

acc. = (mg sin α cos α)/M

So applying the equation, the displacement to be found is:

s = (1/2) [(mg sin α cos α)/M] [2u/(g sin α)]2

= 2mu2/(Mg tan α)

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