Question

Question 6.25 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 (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

Chapter Work, Energy And Power Page 137

Answer 1

Probably this is the answer.
I'm not sure, but I guess this is the way to solve the question.

Answer 2

Two inclined frictionless track AB, and AC are inclined with horizontal ∅1 and ∅2 respectively.
Let mass of stone in AB is m1
and mass of stone in AC is m2

Break the components of all forces by shown in figure :
___________________________

For 1st stone { stone in AB }
potential energy = kinetic energy
m1gh = 1/2m1v1²
v1² = 2gh
v1² = 2 × 10 × 10
v1 = 10√2 m/s
hence, final speed of 1st stone = 10√2m/s
for second stone :
PE = KE
m2gh = 1/2m2v2²
v2² = 2gh
V2 = √2gh = 10√2 m/s
hence, speed of 2nd stone is too 10√2m/s . both have same final speed . e.g speed doesn't depends upon mass of body and inclination of body .

Now,
Let T1 time taken by m1 stone :
see the figure :
net Force on the stone of mass m1 = m1gsin∅1 which is acting downward direction . Let a1 is the acceleration of m1 by which it goes downward .
m1a1 = m2gsin∅1
a1 = gsin∅1
= 10 sin30° = 5 m/s²
now, use formula ,
V = U + at
here, V = 10√2 m/s
u = 0
10√2 = 5T1
T1 = 2√2 sec

similarly for 2nd stone :
Let T2 time taken by 2nd stone :
see figure , Let in a2 acceleration stone moves in downward .
m2a2 = m2gsin∅2
a2 = gsin∅2 = 10×sin60°
= 5√3 m/s
use , formula
V = u + at
10√2 = 5√3T2
T2 = (2√2/√3 ) sec
you can see that ,
T1 > T2
hence , 1st stone takes more time then 2nd stone .