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Grade: 11
        
Give me answers. I have a problem of this question
 
 
 
Fast .  Physics .    Centre of Mass
 
 
 
17 days ago

Answers : (1)

Vikas TU
6930 Points
							
From momenta conservatio in x direction:
Mv2 = mv1cos(thetha)..........(1)
 
From energy conservation,
mgh = 0.5mv1^2 + 0.5Mv2^2...........(2)
Solving both the eqns. we get,
v2 = sqroot(2mgh/((M/cos(thetha))^2 + M)
and 
v1 = (M*sqroot(2mgh/((M/cos(thetha))^2 + M))/(m*cos(thetha))
 
time for block m to get down to the height h would be:
v1 = 0 + gt
t = v1/g
Displacement of wedge will be in x direction only.
Therfore,
x = v2*t + 0
x = v2*v1/g =>  ((M*sqroot(2mgh/((M/cos(thetha))^2 + M))/(m*cos(thetha))*sqroot(2mgh/((M/cos(thetha))^2 + M))/g 
 afer solving ,
x = 2h/(M/(m*cos(thetha)) + costhetha) m.
 
12 days ago
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