A particle is placed at the point A of a frictionless track A B C as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is :
(Take g=10 m / s²).

Solution

Physics sir · 3 steps

No friction, so only the height drop from A to B decides the speed: 1 m down to 0.5 m.

1. Energy is conserved. The track is frictionless and the normal force does no work, so kinetic plus potential energy stays the same. The particle starts at A with almost zero speed because it is only gently pushed.
2. Height actually lost. A is at 1 m and B is at 0.5 m above the bottom line. So the particle drops only h = 1 - 0.5 = 0.5 m. The shape of the curve does not matter.

   1/2mv² = mgh ⇒ v = √2gh = √(2 × 10 × 0.5)

3. So. v = √10 m/s, which is option (c).

The idea to keep. On a frictionless track, speed depends only on the vertical drop between the two points, v = √2gh. Read the heights carefully and take the difference.

Answer from the NTA final answer key. Solution by PrepWiser.