A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Question 17

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Fig. 4.1 (not to scale)

 

The spring is supported and stands vertically. The ball has a speed of 2.8 m s^-1 as it makes contact with the spring. The ball is brought to rest as the spring is compressed.

 

(a) Show that the kinetic energy of the ball as it makes contact with the spring is 0.16 J. [2]

 

 

(b) The variation of the force F acting on the spring with the compression x of the spring is shown in Fig. 4.2.

Fig. 4.2

 

The ball produces a maximum compression XB when it comes to rest. The spring has a spring constant of 800 N m^-1.

 

Use Fig. 4.2 to

(i) calculate the compression XB, [2]

 

(ii) show that not all the kinetic energy in (a) is converted into elastic potential energy in the spring. [2]

 

 

 

 

 

Reference: Past Exam Paper – June 2015 Paper 21 Q4

 

 

 

 

 

Solution:

(a)

kinetic energy = ½ mv²            

kinetic energy = ½ × 0.040 × (2.8)² = 0.157 J or 0.16 J

 

 

(b)

(i)

{From Hooke’s law,}

k = F / x                       or F = kx        

 

{x = F / k}

XB = 14 / 800

XB = 0.0175 m                                    

 

(ii)

{The elastic potential energy stored in the spring can be obtained from the area under the F-x graph.}

 

area under graph = elastic potential energy stored

or Elastic PE = ½ kx²                 or ½ Fx

 

{Elastic PE = ½ Fx = ½ × 14 × 0.0175}

(energy stored =) 0.1225 J less than KE (of 0.16 J)

 

{This value of elastic PE is less than the KE calculated above.}