As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s²)
The value of 'x' to the nearest integer is __________.

Solution

Physics sir · 4 steps

Only the 5 m drop from A to B matters: lost potential energy becomes kinetic energy at B.

1. Set up energy. The particle starts from rest at A, since it is only slightly displaced. The track is taken as smooth, so mechanical energy is conserved between A and B.
2. Height actually lost. A is 10 m above the ground and B is 5 m above it, so the particle falls through h = 10 - 5 = 5 m. The shape of the curve does not matter, only the difference in height.

   mgh = 1/2mv² ⇒ v = √2gh

3. Put in numbers. The mass cancels, so 10 kg is not needed. v = √(2 × 10 × 5) = √100 = 10 m s⁻¹.
4. So. The speed at B is 10 m s⁻¹, so x = 10.

The common slip: 14. Using the full 10 m height of A instead of the 5 m drop to B gives √200 ≈ 14, not 10.

The idea to keep. On a smooth track, speed depends only on how far the body has dropped vertically, v = √2gh. The path shape and the mass do not matter.

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