A car of mass `1000kg` negotiates a banked curve of radius `90m` on a frictionless road. If the banking angle is `45^(@)` the speed of the car is:

Options

A. 1. 20 `ms^(-1)`
B. 2. 30 `ms^(-1)`
C. 3. 5 `ms^(-1)`
D. 4. 10 `ms^(-1)`

Answer

To solve the problem of a car negotiating a banked curve, we can follow these steps:

### Step 1: Understand the Forces Acting on the Car
When the car is on a banked curve, there are two main forces acting on it:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The normal force acting perpendicular to the surface of the banked road.

### Step 2: Draw the Free Body Diagram
In the free body diagram:
- The weight \( mg \) acts vertically downwards.
- The normal force \( N \) acts perpendicular to the inclined surface.
- The car experiences a centripetal force \( F_c \) directed towards the center of the circular path.

### Step 3: Resolve Forces
For the car to negotiate the curve without skidding, the components of the forces must provide the necessary centripetal force:
- The component of the gravitational force acting down the slope: \( F_{g,\text{parallel}} = mg \sin \theta \)
- The component of the normal force providing the centripetal force: \( F_{N,\text{horizontal}} = N \cos \theta \)

### Step 4: Set Up the Equations
Since the road is frictionless, the centripetal force required to keep the car moving in a circle is provided by the horizontal component of the normal force:
\[ N \cos \theta = \frac{mv^2}{r} \]
And the vertical forces must balance:
\[ mg \sin \theta = N \sin \theta \]

### Step 5: Solve for Normal Force
From the vertical force balance, we can express \( N \):
\[ N = \frac{mg \sin \theta}{\sin \theta} = mg \]

### Step 6: Substitute into the Centripetal Force Equation
Now substitute \( N \) into the centripetal force equation:
\[ mg \cos \theta = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides:
\[ g \cos \theta = \frac{v^2}{r} \]

### Step 7: Solve for Speed \( v \)
Rearranging gives us:
\[ v^2 = g \cdot r \cdot \cos \theta \]
Substituting \( g = 10 \, \text{m/s}^2 \), \( r = 90 \, \text{m} \), and \( \theta = 45^\circ \) (where \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)):
\[ v^2 = 10 \cdot 90 \cdot \frac{1}{\sqrt{2}} \]
Calculating this gives:
\[ v^2 = 900 \cdot \frac{1}{\sqrt{2}} = 450\sqrt{2} \]
Taking the square root:
\[ v = \sqrt{450\sqrt{2}} = 30 \, \text{m/s} \]

### Final Answer
The speed of the car is \( 30 \, \text{m/s} \).

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