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When dealing with projectile motion on an inclined plane, it's critical to understand how the initial velocity of the projectile breaks down into components. Given that a particle is thrown with a velocity of 10 m/s at a 60° angle to the horizontal, this velocity can be split into two main parts using trigonometry.

• The horizontal component is found using: \( v_{0x} = v_0 \cos(60°) = 10 \times \frac{1}{2} = 5 \ \mathrm{ms}^{-1} \)
• The vertical component is: \( v_{0y} = v_0 \sin(60°) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \ \mathrm{ms}^{-1} \)

These components are essential because they allow us to analyze the motion of the particle separately in the horizontal and vertical directions. This decomposition helps in understanding how the projectile behaves as it travels along the incline.

The physics of an inclined plane involves understanding how gravity affects objects on an inclined surface. When a projectile moves along an inclined plane, gravity acts in two directions relative to the incline: parallel and perpendicular.

• The gravitational component parallel to the incline is given by: \( a_{\parallel} = g \sin(30°) = \frac{g}{2} \)
• The component perpendicular to the incline is: \( a_{\perp} = g \cos(30°) = \frac{\sqrt{3}g}{2} \)

Understanding these components allows us to determine how the gravity-induced acceleration affects the projectile's motion along and perpendicular to the inclined plane. When the direction of the projectile's velocity becomes parallel to the incline, it’s crucial to compute the interaction between these components accurately.

Decomposing the forces acting on a projectile, especially when on an incline, is crucial for analyzing its motion. This involves breaking down both the velocity and acceleration into components which are easier to handle mathematically.To determine when the velocity becomes parallel to the incline, we set the equation for parallel velocity to zero, which reflects that the perpendicular influence has been neutralized. Using vector decomposition helps simplify the complex interactions at play:

• Set the equation for velocity parallel to the incline: \( v_{\parallel}(t) = v_{0x}(\cos 30^{\circ}) + v_{0y}(\sin 30^{\circ}) - \frac{g}{2}t = 0 \)
• Substituting known values, we solve for time: \( t = \frac{2}{\sqrt{3}} \ \mathrm{s} \)

This solution tells us that after \( \frac{2}{\sqrt{3}} \) seconds, the projectile's velocity becomes parallel to the incline. Understanding how to decompose forces and using the correct trigonometric identities is key to solving such problems effectively.