Question

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

• A. $ 20\,ms^{-1} $
• B. $ 30 \,ms^{-1} $
• C. $ 5\, ms^{-1} $
• D. $ 10\, ms^{-1} $

Answer 1

The Correct Option is B

To solve the problem of determining the speed of the car on a frictionless banked curve, we use the concept of circular motion on a banked road. When a car travels on a banked curve, the necessary centripetal force is provided by the horizontal component of the normal force due to the banking.

For a banked curve with an angle $\theta$, radius $r$, and with gravity $g = 9.8 \, ms^{-2}$, the optimum speed $v$ of the car that doesn't rely on friction is given by:

$$ v = \sqrt{rg \cdot \tan(\theta)} $$

Given:

• Mass of car, $m = 1000\, kg$ (not needed in a frictionless scenario).
• Radius of curve, $r = 90\, m$
• Banking angle, $\theta = 45^{\circ}$

The tangent of the angle $\theta = 45^{\circ}$ is:

$$ \tan(45^{\circ}) = 1 $$

Substituting the given values into the formula:

$$ v = \sqrt{90 \times 9.8 \times 1} $$

Calculate the value:

$$ v = \sqrt{882} $$

Upon calculation, we get:

$$ v \approx 29.7\, ms^{-1} $$

Rounding this to the nearest whole number, we find:

$$ v \approx 30\, ms^{-1} $$

Thus, the correct speed of the car negotiating the banked curve is $30 \,ms^{-1}$.

Conclusion: The correct answer is $30 \,ms^{-1}$. This matches the given correct answer.