IIT JEE , AIEEE Forum - Physics , Mechanics

akhil_o

mgh=1/2mv²
v=

2gh

let the impact force be F
drawing FBD we can see that for the sphere,
two forces of eqal magnitude(F) act on it at angle 30 deg
so we have

2Fsin 30=(mv₁+m

2gh)/

t
or
F

t=mv₁+m

2gh---------(i)
for one wedge
F sin 60=Mv₂

t due to change in momentum
F

3/2

t=Mv_{2-----------------(ii)}

from (i)a dn (ii) we have

2/

3 Mv₂=mv₁+mv₀

v₀cos 60 e=v₂cos 30+v₁cos 60

putting values
ev₀=v₁+

3 v₂

substitute these values in above equations
v₂=(

3)(1+e) m(

2gh)/2M+3m

v₁=(2eM-3m)(

2gh)/2M+3m

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