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.

Question 9

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]

Reference: Past Exam Paper – November 2010 Paper 23 Q4

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