Dynamics Problem Solution: Projectile Motion and Impact Analysis

Core Principles and Given Parameters (Dynamics)

• Subject Context: The problem involves the principles of kinematics and kinetics of a particle, specifically looking at projectile motion combined with coefficients of restitution during impact with a fixed surface.
• Initial Conditions at Point A (Throwing Point):
  • Mass of the Ball ($m$): $0.5\,\text{kg}$
  • Initial Velocity ($v_A$): $10\,\text{m/s}$
  • Angle of Release ($\theta$): $30^\circ$ above the horizontal.
  • Initial Height ($y_0$): $1.5\,\text{m}$
  • Horizontal Distance to Wall ($d$): $3\,\text{m}$
• Environmental and Material Constants:
  • Acceleration due to Gravity ($g$): Standard Earth gravity is used, assumed as $9.81\,\text{m/s}^2$.
  • Coefficient of Restitution ($e$): $0.5$. This value dictates the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact.

Kinematics of Initial Motion (A to B)

• Decomposition of Initial Velocity:
  • The horizontal component of the velocity ($v_{Ax}$) remains constant throughout the flight (neglecting air resistance).
    • $v_{Ax} = v_A \times \cos(30^\circ) = 10\,\text{m/s} \times \cos(30^\circ)$
    • $v_{Ax} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.6603\,\text{m/s}$
  • The vertical component of the velocity ($v_{Ay}$) describes the initial upward motion.
    • $v_{Ay} = v_A \times \sin(30^\circ) = 10\,\text{m/s} \times \sin(30^\circ)$
    • $v_{Ay} = 10 \times 0.5 = 5\,\text{m/s}$
• Time of Flight to the Wall ($t_B$):
  • Using the horizontal distance equation: $x = v_{Ax} \times t$
    • $3\,\text{m} = 8.6603\,\text{m/s} \times t_B$
    • $t_B = \frac{3}{8.6603} \approx 0.3464\,\text{s}$

Impact Analysis at Point B (Striking Velocity)

• Horizontal Velocity Component at B ($v_{Bx}$):
  • There are no horizontal forces; thus, the velocity is unchanged.
  • $v_{Bx} = v_{Ax} = 8.6603\,\text{m/s}$
• Vertical Velocity Component at B ($v_{By}$):
  • Using the kinematic equation: $v_y = v_{y0} - g \times t$
    • $v_{By} = 5\,\text{m/s} - (9.81\,\text{m/s}^2 \times 0.3464\,\text{s})$
    • $v_{By} = 5 - 3.3982 = 1.6018\,\text{m/s}$
  • The positive sign indicates the ball is still moving upward when it strikes the wall.
• Magnitude and Direction of Strike Velocity ($v_B$):
  • Magnitude:
    • $v_B = \sqrt{v_{Bx}^2 + v_{By}^2}$
    • $v_B = \sqrt{(8.6603)^2 + (1.6018)^2} = \sqrt{75 + 2.5658} \approx \sqrt{77.5658} \approx 8.807\,\text{m/s}$
  • Direction ($\phi$):
    • $\phi = \arctan\left(\frac{v_{By}}{v_{Bx}}\right) = \arctan\left(\frac{1.6018}{8.6603}\right)$
    • $\phi \approx \arctan(0.1849) \approx 10.48^\circ$ above the horizontal.

Rebound Dynamics and Coefficient of Restitution

• Impact Modeling:
  • The wall is perfectly vertical; therefore, the normal to the surface of impact is the horizontal axis ($x$-axis).
  • The parallel direction is the vertical axis ($y$-axis).
• Post-Rebound Velocity Components ($v'_{Bx}$, $v'_{By}$):
  • Horizontal Component (Normal to Impact):
    • Applying the coefficient of restitution $e$:
    • $v'_{Bx} = -e \times v_{Bx}$
    • $v'_{Bx} = -0.5 \times 8.6603\,\text{m/s} = -4.3302\,\text{m/s}$
    • (The negative sign indicates the direction reversed, moving away from the wall).
  • Vertical Component (Parallel to Impact):
    • Assuming a smooth wall, friction is ignored, and momentum in the parallel direction is conserved.
    • $v'_{By} = v_{By} = 1.6018\,\text{m/s}$
• Magnitude and Direction of Rebound Velocity ($v'_B$):
  • Magnitude:
    • $v'_B = \sqrt{(v'_{Bx})^2 + (v'_{By})^2}$
    • $v'_B = \sqrt{(-4.3302)^2 + (1.6018)^2} = \sqrt{18.7506 + 2.5658} \approx \sqrt{21.3164} \approx 4.617\,\text{m/s}$
  • Direction ($\gamma$):
    • $\gamma = \arctan\left(\frac{v'_{By}}{|v'_{Bx}|}\right) = \arctan\left(\frac{1.6018}{4.3302}\right)$
    • $\gamma \approx \arctan(0.3699) \approx 20.30^\circ$ above the horizontal (moving away from the wall).

Trajectory Post-Impact (B to C)

• Determining the Vertical Position of Point B ($y_B$):
  • The vertical height attained at the moment of impact with the wall.
  • $y_B = y_0 + v_{Ay} \times t_B - \frac{1}{2} \times g \times t_B^2$
  • $y_B = 1.5 + (5 \times 0.3464) - (0.5 \times 9.81 \times 0.3464^2)$
  • $y_B = 1.5 + 1.732 - 0.5886 = 2.6434\,\text{m}$
• Time Calculation for Motion from B to Ground (Point C):
  • The ball falls from height $y_B$ to $y_C = 0$.
  • The vertical equation for the rebound trajectory:
    • $y_C = y_B + v'_{By} \times t_{BC} - \frac{1}{2} \times g \times t_{BC}^2$
    • $0 = 2.6434 + 1.6018 \times t_{BC} - 4.905 \times t_{BC}^2$
  • Rearranging into standard quadratic form: $4.905 \times t_{BC}^2 - 1.6018 \times t_{BC} - 2.6434 = 0$:
    • Applying quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
    • $t_{BC} = \frac{1.6018 + \sqrt{(-1.6018)^2 - 4(4.905)(-2.6434)}}{2(4.905)}$
    • $t_{BC} = \frac{1.6018 + \sqrt{2.5658 + 51.8633}}{9.81}$
    • $t_{BC} = \frac{1.6018 + \sqrt{54.4291}}{9.81} = \frac{1.6018 + 7.3776}{9.81} \approx 0.9153\,\text{s}$
• Horizontal Distance from Wall ($s_{BC}$):
  • The distance traveled horizontally away from the wall until striking point C.
  • $s_{BC} = |v'_{Bx}| \times t_{BC}$
  • $s_{BC} = 4.3302\,\text{m/s} \times 0.9153\,\text{s} \approx 3.963\,\text{m}$

Summary of Results

• Part A (Strike Velocity):
  • Magnitude: $8.81\,\text{m/s}$
  • Direction: $10.5^\circ$ above horizontal.
• Part B (Rebound Velocity):
  • Magnitude: $4.62\,\text{m/s}$
  • Direction: $20.3^\circ$ above horizontal (away from wall).
• Part C (Distance to ground strike):
  • The ball strikes the ground at point C, located $3.96\,\text{m}$ from the base of the wall.