A car of mass $1000 ext{ kg}$ negotiates a banked curve of radius $90 ext{ m}$ on a frictionless road. If

Question

A car of mass $1000 	ext{ kg}$ negotiates a banked curve of radius $90 	ext{ m}$ on a frictionless road. If the banking angle is $45^{\circ}$, the speed of the car is:

Accepted Answer

1. **Identify the Condition:** The car travels on a banked road where friction is negligible (frictionless). The necessary centripetal force is provided solely by the horizontal component of the normal reaction force.
2. **Formula:** For a vehicle moving on a frictionless banked curve, the optimum speed $v$ is given by the relation:
   $$ v = \sqrt{Rg 	an 	heta} $$
   where $R$ is the radius of curvature, $g$ is the acceleration due to gravity, and $	heta$ is the angle of banking [NCERT Physics Class 11, Laws of Motion, Section 4.10, Eq. 4.22].
3. **Substitution:** Given $R = 90 	ext{ m}$, $	heta = 45^{\circ}$, and taking $g = 10 	ext{ m s}^{-2}$ (standard approximation for this calculation):
   $$ v = \sqrt{90 	imes 10 	imes 	an 45^{\circ}} $$
4. **Calculation:** Since $	an 45^{\circ} = 1$:
   $$ v = \sqrt{900 	imes 1} = 30 	ext{ m s}^{-1} $$
   (Note: The mass of the car is not required for this calculation).

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