Physics Mechanics

1. A wedge of mass M lies on a smooth horizontal plane while a block of mass m=M/2 is in contact with the smooth inclined face of the wedge . The inclination of this face to the horizontal is 30°. The system is released from rest with the block at a vertical height H above the horizontal plane. When the block reaches the horizontal plane, find (i) the speed of the wedge, and (ii) the distance the wedge has travelled.


My approach is like this:

Normal reaction acting on the wedge by block

= Force acting on the block

= mg cos30

By basic geometry, horizontal component of the force= sin30 mg cos30

Acceleration= 1/2(mgcos30)/(M)=1/2(mgcos30)/(2m)

Update 2:

Considering the motion of the block:

s=0.5 at^2+ut



t= 2sqrt(2H/g)

By v=u+at,



=1.918sqrtH m/s

Update 3:

By s=0.5at^2+ut

distance= 0.5(1/4)gcos30 (8H/g)

= gcos30H/g


=0.866H m

Anyone can tell me whether I am correct or not?

Thanks a lot!

1 Answer

  • 天同
    Lv 7
    6 years ago
    Favorite Answer

    Consider the block of mass m, acceleration of block down the wedge = g.sin(30)

    Use equation of motion: s = ut + (1/2)at^2

    with s = H/sin(30), u = 0 m/s, a = g.sin(30) m/s^2, t=? (where t is the time the block reaches the horizontal plane)

    hence, H/sin(30) = (1/2).(g.sin(30))t^2 (take g = 10 m/s^2)

    t = 0.8944[sqrt(H)] where "sqrt" stands for "square-root"

    Consider the wedge, force acting onto the inclined face = mg.cos(30)

    Hence, horizontal force pushing the wedge to move

    = [mg.cos(30)].sin(30)

    Horizontal acceleration of wedge = mg.cos(30).sin(30)/M = g.cos(30).sin(30)/2

    = 0.2165 m/s^2 (take g = 10 m/s^2)

    (i) Speed of wedge when the block reaches the plane

    = 0.2165 x 0.8944(sqrt(H)) = 0.1936.sqrt(H)

    (ii) Distance travelled by the wedge

    = (1/2).(0.2165).[0.8944sqrt(H)]^2 (use s = ut + (1/2)at^2 )

    = 0.0866H

    2014-09-05 19:49:19 補充:

    sorry, I made a wrong numerical calculation in the wedge acceleration. It should be,

    Horizontal acceleration of wedge = mg.cos(30).sin(30)/M = g.cos(30).sin(30)/2

    = 2.165 m/s^2 (take g = 10 m/s^2)

    2014-09-05 19:49:43 補充:

    (cont'd)... Hence, the speed in (i) and distance in (ii) should be 10 times the given figures, i.e. 1.936sqrt(H) and 0.866H respectively.

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