Question 5 (paper Question V) — Ball vs. Wall Impact and Rebound Trajectory (Part B · Dynamics, equal value): 04-BS-3, Undated paper, Q5

Question 5 (paper Question V) — Ball vs. Wall Impact and Rebound Trajectory (Part B · Dynamics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ball thrown from A, height 1.5 m, $v_A=12$ m/s at $30^\circ$ above horizontal, toward a wall 3 m away. Coefficient of restitution at the wall $e=0.5$.

Find. (a) velocity at B (wall impact); (b) rebound velocity; (c) distance $s$ from the wall to landing point C.

Figure 5. Ball thrown at 12 m/s, 30° above horizontal from 1.5 m, strikes wall at B, rebounds to land at C.

Approach. Projectile motion (constant $v_x$, $v_y$ under gravity) to the wall; the wall reverses and scales the horizontal (normal) velocity component by $e$ and leaves the vertical (tangential) component unchanged; then a second projectile-motion segment to the ground.

Launch components. $v_{Ax}=12\cos30^\circ=10.392$ m/s, $v_{Ay}=12\sin30^\circ=6.000$ m/s.
Time to reach the wall ($x=3$ m): $t_1=3/10.392=0.2887$ s.
Velocity and height at B. $v_{Bx}=10.392$ m/s (unchanged); $v_{By}=6.000-9.81(0.2887)=3.168$ m/s (still rising). $\boxed{|\vec v_B|=\sqrt{10.392^2+3.168^2}=10.86\text{ m/s},\ 16.9^\circ\text{ above horizontal}}$. Height: $y_B=1.5+6.0(0.2887)-4.905(0.2887)^2=\boxed{2.823\text{ m}}$.
Rebound at the wall. $v'_{Bx}=-e\,v_{Bx}=-0.5(10.392)=\boxed{-5.196\text{ m/s}}$ (away from wall); $v'_{By}=v_{By}=\boxed{3.168\text{ m/s}}$ (unchanged, tangential to the wall).
Time to fall from B to the ground. $2.823+3.168t_2-4.905t_2^2=0\Rightarrow t_2=1.1475$ s (positive root).
Distance $s$. $s=|v'_{Bx}|\,t_2=5.196(1.1475)=\boxed{5.96\text{ m}}$ (measured from the wall, per Figure 5).

Final Results — Question 5
Quantity	Value
(a) Velocity striking wall at B	10.86 m/s, 16.9° above horizontal
(b) Rebound velocity from wall	$(-5.20,\ 3.17)$ m/s — 6.09 m/s, 31.4° above horizontal
(c) Distance $s$ (wall to C)	5.96 m

Topic: Projectile motion combined with impact (coefficient of restitution) against a fixed vertical surface.

Key relations: constant-acceleration kinematics $v=v_0+at$, $y=y_0+v_0t+\tfrac12at^2$; restitution $e=-\dfrac{v'_{\text{normal}}}{v_{\text{normal}}}$ applied to the component perpendicular to the impacted surface only.

Why this works. A rigid, immovable vertical wall exerts an impulsive force purely horizontally, so only the horizontal (normal) velocity component is affected by $e$; the vertical (tangential) component, parallel to the wall, is unchanged through the impact — this splits one 2D impact problem into two independent 1D calculations.

Common pitfall: applying $e$ to the vertical component instead of the horizontal one (the wall is vertical, so the horizontal direction is the impact normal); also forgetting the ball is still rising when it reaches the wall (check $t_1$ against time-to-apex $=v_{Ay}/g=0.612$ s).

Source: Hibbeler, Engineering Mechanics: Dynamics, Ch. 12 (Kinematics of a Particle — projectile motion) and Ch. 15 (Impact).

Question 5 of 6: Question 5 (paper Question V) — Ball vs. Wall Impact and Rebound Trajectory (Part B · Dynamics, equal value)

Question 5 (paper Question V) — Ball vs. Wall Impact and Rebound Trajectory (Part B · Dynamics, equal value)