Physics Mechanics

2014-09-05 11:22 pm

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.

更新1:

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)

更新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

更新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!

回答 (1)

天同

2014-09-06 12:15 am

✔ 最佳答案

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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