question_answer

Two inclined frictionless tracks, one gradual and the other steep meet at point A from where two stones (I and II) are allowed to slide down from rest, one on each track as shown in figure. Which of the following statements is correct?

A) Both the stones reach the bottom at the same time but not with the same speed.

B) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.

C) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.

D) Both the stones reach the bottom at different times and with different speeds.

Correct Answer: C

Solution :

AB and AC are two smooth planes inclined to the horizontal at different angles. As height of both the planes are same, therefore, both the stones will reach the bottom with same speed. \[\because v=\sqrt{2gh}\] \[\therefore {{v}_{1}}={{v}_{2}}\] From figure, acceleration, \[{{a}_{1}}=g\text{ }sin\text{ }{{\theta }_{1}}\] acceleration, \[{{a}_{2}}=g\text{ }sin\text{ }{{\theta }_{2}}\] As \[{{\theta }_{2}}>{{\theta }_{1}},\] so \[{{a}_{2}}>{{a}_{1}}\] Using, \[v=u+at\] \[{{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}};{{v}_{1}}={{a}_{1}}{{t}_{1}}\] \[{{t}_{1}}=\frac{{{V}_{1}}}{{{a}_{1}}}\] Similarly, \[{{t}_{2}}=\frac{{{V}_{2}}}{{{a}_{2}}}\because {{a}_{2}}>{{a}_{1}}\] So, \[{{t}_{2}}<{{t}_{1}}\] Hence, stone II takes less time to reach the bottom.