1. Motion Analysis

Given: $u = 5.0$ m/s, $h = 0.8$ m, $g = 10$ m/s². Velocity of ball at the top of the box ($v_{top}$): $$ v_{top}^2 = u^2 – 2gh = 25 – 2(10)(0.8) = 25 – 16 = 9 $$ $$ v_{top} = 3.0 \text{ m/s} $$ Time to go up ($t_{up}$) = $\frac{u – v_{top}}{g} = \frac{5 – 3}{10} = 0.2$ s. Total time period for one round trip (up and down) $T = 2 \times 0.2 = 0.4$ s.

2. Average Force on Top ($F_T$)

The ball hits the ceiling with $v_{top} = 3$ m/s and rebounds elastically ($-3$ m/s). Impulse imparted to the ceiling (upward force on box): $$ J_T = m(v_{top} – (-v_{top})) = 2mv_{top} = 2(1.0)(3.0) = 6.0 \text{ Ns} $$ Average Force: $$ F_T = \frac{J_T}{T} = \frac{6.0}{0.4} = 15 \text{ N} $$ Direction: Upward.

3. Average Force on Bottom ($F_B$)

The ball hits the floor with $u = 5$ m/s and rebounds ($5$ m/s). Impulse imparted to the floor (downward force on box): $$ J_B = m(u – (-u)) = 2mu = 2(1.0)(5.0) = 10.0 \text{ Ns} $$ Average Force: $$ F_B = \frac{J_B}{T} = \frac{10.0}{0.4} = 25 \text{ N} $$ Direction: Downward.

4. Weighing Machine Reading

The weighing machine measures the total downward force exerted by the box. $$ W = M_{box}g + F_{avg, net} $$ Net average force from ball on box = $F_B$ (down) – $F_T$ (up). $$ F_{ball \to box} = 25 – 15 = 10 \text{ N} $$ Weight of box = $2.0 \times 10 = 20$ N. Total Reading $W = 20 + 10 = 30$ N.

Check: Treat the box + ball as a system. The center of mass of the ball does not accelerate on average over a full cycle (it oscillates). Thus, the average external force must equal the total weight. $W = (M+m)g = (2.0 + 1.0) \times 10 = 30$ N.