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)?

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Answer 1 · 2018-01-05T18:38:53

10 m

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)#