第16題和第17題?

16. Consider the 1st (at position 0 cm) and 2nd (at position 0.5 cm) images.
Use equation of motion: s = ut + (1/2)at^2
with s = 0.5 cm, t = 0.1 s,
hence, 0.5 = u(0.1) + (1/2)a(0.1)^2
i.e. 0.5 = 0.1u + 0.005a
u = 5 - 0.05a -------- (1)

Consider the 1st (at 0 cm) and 3rd (at position 1.5 cm) images.
Use the same equation as above, with s = 1.5 cm, t = 0.2 s
hence, 1.5 = u(0.2) + (1/2)a(0.2)^2 ---------- (2)
substitute u from (1) into (2),
1.5 = (5 - 0.05a)(0.2) + 0.02a
solve for a gives a = 50 cm/s^2 = 0.5 cm/s^2

Given the mass of the ball is 100 g = 0.1 kg
hence, external force = 0.1 x 0.5 N = 0.05 N

The answer is option A.

17. The diagram is missing.