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? 
The given question can be illustrated using the figure below: AB and AC are two smooth planes inclined to the horizontal at ∠θ1 and ∠θ2 respectively. The height of both the planes is the same, therefore, both the stones will reach the bottom with same speed. As P.E. at O = K.E. at A = K.E. at B Therefore, mgh = 1/2 mv12 = 1/2 mv22 ∴ v1 = v2 As it is clear from fig. above, acceleration of the two blocks are a1 = g sin θ1 a2 = g sin θ2As θ2 > θ1 ∴ a2 > a1 From v = u + atat As t ∝ 1/a, and a2 > a1 ∴ t2 < t1 That is, the second stone will take lesser time and reach the bottom earlier than the first stone.
Option 3 : greater momentum
10 Qs. 10 Marks 10 Mins
CONCEPT:
The expression for kinetic energy is given by: Where m = mass of the body and v = velocity of the body
Momentum (p) = mass (m) × velocity (v) The relationship between the kinetic energy and Linear momentum is given by: As we know, Divide numerator and denominator by m, we get [p = mv] EXPLANATION: Given that: K.E1 = K.E2= K.E (let say) The relation between the momentum and the kinetic energy is given by: But as K.E is same ∴ Or, Here, if m1 > m2 then p1 > p2
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