3.1 Define the term free fall - NSC Physical Sciences - Question 3 - 2023 - Paper 1

To calculate the initial speed of the ball ($v_0$) when projected upwards, we can use the following equation of motion:

$v^2 = u^2 + 2a s$

where:

• $v$ is the final velocity (0 m/s at the peak),
• $u$ is the initial velocity,
• $a$ is the acceleration (-9.81 m/s², acting downward), and
• $s$ is the displacement (5.89 m).

Substituting the values:

$0 = u^2 - 2(9.81)(5.89)$

Solving for $u$ gives:

$u^2 = 2 imes 9.81 imes 5.89$ $u^2 = 115.854$ $u = ext{sqrt}(115.854) \\ u ≈ 10.77 \, m/s$

The kinetic energy before the collision can be calculated using the formula:

$KE_i = \frac{1}{2}mv^2$

where:

• $m = 0.5 \, kg$ and $v = 11.92 \, m/s$.

Calculating:

$KE_i = \frac{1}{2} (0.5) (11.92)^2 = \frac{1}{2} (0.5) (141.7984) = 35.45 \, J$

For the kinetic energy after the collision (just before hitting the ground):

$KE_f = \frac{1}{2} m v^2 = \frac{1}{2} (0.5) (0)^2 = 0 \, J$

Thus, the kinetic energy lost is:

$\Delta KE = KE_i - KE_f = 35.45 - 0 = 35.45 \, J$

To find the time taken ($t$) for the ball to reach point P after being projected upwards, we can use the formula:

$v = u + at$

where:

• $v = 0$ (velocity at the peak),
• $u = 11.92 \, m/s$,
• $a = -9.81 \, m/s^2$.

Rearranging gives:

$0 = 11.92 - 9.81t \\ 9.81t = 11.92 \\ t = \frac{11.92}{9.81} \\ t ≈ 1.21 \, s$

Based on the provided graph:

• 3.4.1 K should correspond to the initial velocity after projection, approximately $v = 11.92 \, m/s$.
• 3.4.2 L should be the maximum velocity (in terms of time and peak height) just before the ball starts its descent, which should be 0 m/s (at its peak).
• 3.4.3 The time difference $t2 - t1$ corresponds to the time taken to go up and come back down to point P, calculated earlier as $t$ times 2, thus approximately $2 imes 1.21 \, s = 2.42 \, s$.