__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 *X*B 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 *X*B,
[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 mv^{2}

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 kx^{2} 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.}