Two inclined frictionless tracks , one gradual and the other steep meet at A from where two stones are allowed to slide down from rest , one on each track as shown in figure . Will the stones reach there with the same speed ? Explain . Given `theta_(1) = 30^(@) , theta_(2) = 60^(@) and h = 10 m ` What are the speeds and times taken by the two stones ?

Explanation

No suppose at time `t_(1)` reached at ground
` :. ` From equation of motion
`v= v_(0)+"gt_(1),v_(0)= 0 and g = gc cos 60^(@)`
` :. " "t_(1)=v/(gxx1/2)=(2xx14.14)/10=(2xxsqrt(2)xx10)/10=2sqrt(2)s`
Suppose at time `t_(2)` ball - 2 reaches at ground ,
` :. "In" v = v_(0)+"gt"_(2),v_(0),g = g cos^(@)`
` :. v = g cos 30^(@)t_(2)`
` :. t_(2)=v/(g cos 30^(@))=(2sqrt(2)xx10)/(10xxsqrt(3)) =(2sqrt(2))/(sqrt(3))`
Both balls having time to reach at ground and hence they cannot reached together . Yes , bot balls are released drom height h = 10 cm .
potential energy at height h = kinetic energy at bottom .
` :. mgh = 1/2 mv^(2)`
` :. gh = v^(2)`
` :. v = sqrt(2gh)`
For both ball g and h are unchaned and the speed of balls at the bottom is same .
` :. v = sqrt(10 xx10 xx2)`
` :. v = 10sqrt(2= 10 xx1.414 = 14.14 ms^(-1)`