A 15g ball is shot from a spring gun whose spring has force constant of 600N/m. The spring is compressed by 5cm. What is the greatest possible horizontal range of the ball for this compression is (take g=10m/s^2)?

1 Answer
Jan 5, 2018

10 m

Explanation:

The energy stored in the spring is given by:

sf(E=1/2kx^2)

k is the force constant.

x is the extension.

:.sf(E=1/2xx600xx0.05^2=0.75color(white)(x)J)

I will assume that all this energy will appear as the kinetic energy of the ball:

:.sf(KE=1/2mv^2)

sf(v=sqrt((2KE)/m)=sqrt((2xx0.75)/0.015)=10color(white)(x)"m/s")

I will assume that the ball is being launched from the ground as no height is given. The range is given by:

sf(d=(v^2sin2theta)/g)

Where sf(theta) is the angle of launch. To find the value which will give the maximum range we find the 1st derivative and set it to zero.

Using The Chain Rule:

sf((d(d))/(d(theta))=v^2/gxx2cos2theta=0)

:.sf(cos2theta=0)

:.sf(2theta=pi/2)

sf(theta=pi/4=45^@)

This is the launch angle that gives the maximum range.

Using this value gives:

sf(d=(10^2xxsin90)/10=(100xx1)/10=10color(white)(x)m)