THYE asked in 科學及數學其他 - 科學 · 6 years ago

# 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.

Update:

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

H/sin30=1/2(0.5g)(t^2)

H=1/8(gt^2)

t= 2sqrt(2H/g)

By v=u+at,

v=at=1/2(gcos30)(2sqrt(2H/g)

=sqrt(6gH)/4

=1.918sqrtH m/s

Update 3:

By s=0.5at^2+ut

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

= gcos30H/g

=cos30H

=0.866H m

Anyone can tell me whether I am correct or not?

Thanks a lot!

Rating
• 天同
Lv 7
6 years ago

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.