Question (6): (6 marks)
In the conservation of mechanical energy experiment, the following results were obtained:
The mass of the plastic ball m = 25 g, the distance between the two photogates at position B is $l$ = 0.10
m, the height of the top of the table from the floor H = 75 cm, as shown; $t_1$ is time between the two
photogates, and $t_2$ is the time between the second photogate and the timer plate at C.
i- Complete the following table
$t_1$
(s)
$t_2$
(s)
$V_{Cx}$ (m/s) $V_{Cy}$ (m/s)
Mechanical
energy at B
$E_B$ (J)
Mechanical
energy at C
$E_C$ (J)
0.045 0.391 2.2 1.49 0.578 0.578
$V_{Cx} = V_B = \frac{l}{t_1} = \frac{0.10}{0.045} = 2.2 \frac{m}{s}$
$V_{Cy} = gt_2^2 = (9.81)(0.301)^2 = 1.499 \frac{m}{s}$
$V_C = \sqrt{V_{Cx}^2 + V_{Cy}^2} = \sqrt{2.2^2 + 1.49^2} = 6.8 \frac{m}{s}$.
$E_B = \frac{1}{2}mv^2 + mgh$
$E_C = \frac{1}{2}mv^2 + mgh^0 = \frac{1}{2}(\frac{25}{1000})(6.8)^2 = 0.578$
ii- Is the mechanical energy between B and C conserved? Explain
iii- What is meant by conservative force?
The work is independent of Path (gravitational work)
