The Question is from NEET guide , and it was created by the author THIRUPATHI REDDY

Question

A ball is launched with the same speed up two frictionless ramps, one inclined at $30^o$ and the other at $60^o$ to the horizontal.  If the ball travels a distance $x_1$ up the $60^o$ ramp and $x_2$ up the $30^o$ ramp, what is the relationship between $x_1$ and $x_2$?

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Work-Energy Theorem, Inclined Planes, Kinematics
	**Prerequisites**: Basic trigonometry, Energy principles
## SOLUTION PATHWAY
**Given Information**:
	- Same launch speed ($v$) for both ramps.
	- Ramp 1 angle: $60^o$
	- Ramp 2 angle: $30^o$
	- Distance up ramp 1: $x_1$
	- Distance up ramp 2: $x_2$
	- Frictionless ramps (implies no energy loss due to friction).
**Method & Steps**:
	1. **Energy Conservation**: The initial kinetic energy of the ball is converted entirely into potential energy at the highest point on each ramp.
		Reasoning: No friction means mechanical energy is conserved.
		Formula/Rule: $KE_i = PE_f$
	2. **Expressing Energies**: $\frac{1}{2}mv^2 = mgh$, where $h$ is the vertical height gained.
		Working:  For ramp 1: $h_1 = x_1\sin(60^o)$. For ramp 2: $h_2 = x_2\sin(30^o)$.
		Key Point: The height is related to the distance traveled along the ramp by the sine of the inclination angle.
	3. **Relating $x_1$ and $x_2$**:  Since the launch speed $v$ is the same, we have $mgh_1 = mgh_2$, which simplifies to $h_1 = h_2$.
		Result: $x_1\sin(60^o) = x_2\sin(30^o)$ => $x_1(\frac{\sqrt{3}}{2}) = x_2(\frac{1}{2})$
		Verification: Solving for $x_2$ gives $x_2 = \sqrt{3}x_1$.
## OPTION ANALYSIS
[A] x_2 = x_1 :x: This would only be true if the angles were the same.
[B] x_2 = 2x_1 :x: This doesn't account for the trigonometric relationship between distance and height.
[C] x_2 = \sqrt{3} x_1 ✓: This is the correct relationship derived from energy conservation and the geometry of the ramps.
[D] x_2 = 3x_1 :x: This overestimates the relationship between $x_1$ and $x_2$.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly using cosine instead of sine: Remember that the vertical height is related to the sine of the inclination angle.
		2. Forgetting to equate the potential energies: The key is that the initial kinetic energy (and hence the final potential energy) is the same in both cases.
	**Quick Tips**:
		1. Visualize the ramps: A steeper ramp (larger angle) means a smaller distance traveled for the same height gained.
		2. Remember the sine values for common angles: $\sin(30^o) = \frac{1}{2}$, $\sin(60^o) = \frac{\sqrt{3}}{2}$.

THIRUPATHI REDDY · Answer

x_2 = x_1

THIRUPATHI REDDY · Answer

x_2 = 2x_1

THIRUPATHI REDDY · Answer

The correct Option is "x_2 = \sqrt{3} x_1"

THIRUPATHI REDDY · Answer

x_2 = 3x_1

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