Q#9 (Past Exam Paper – November 2010 Paper 23 Q4)
A student takes measurements to determine a value for the acceleration of free fall. Some of the apparatus used is illustrated in Fig. 4.1.
Fig. 4.1
The student measures the vertical distance d between the base of the electromagnet and the bench. The time t for an iron ball to fall from the electromagnet to the bench is also measured.
Corresponding values of t 2 and d are shown in Fig. 4.2.
Fig. 4.2
(a) On Fig. 4.2, draw the line of best fit for the points. [1]
(b) State and explain why there is a non-zero intercept on the graph of Fig. 4.2. [2]
(c) Determine the student’s value for
(i) the diameter of the ball, [1]
(ii) the acceleration of free fall. [3]
Solution:
(a) An acceptable straight line (touching every point) should be drawn
(b)
The distance fallen by the iron ball is not d.
Distance d is the distance fallen by the ball plus the diameter of the ball
{Notice from the diagram that d is NOT measured from the bottom of the ball but from the top. So this distance includes the diameter of the ball}
(c)
(i)
{The diameter is the value of y-intercept. That is, when the ball is touching the ground (at time zero), the value of d is equivalent to the diameter of the ball.}
Diameter of ball: (allow) 1.5 ± 0.5 cm
(ii)
Gradient of graph = 4.76 ± 0.1. The origin {point (0, 0)} should not be used {since the line intercepts the y-axis at a value greater than 0}
{A graph of distance against (time)2 is plotted.
Consider the equation for uniformly accelerated motion:
s = ut + ½at2 where d = s and a = g here.
Compare with y = mx + c where m is the gradient and c the y-intercept.
From equation of motion, gradient of s-t2 graph {s on y-axis and t2 on x-axis} would be equal to ½ g
Gradient of graph = g / 2
So, the acceleration of free fall, g = 9.5 m s-2
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