A) 6.0m
B) 10.0m
C) 12.0m
D) 8.0m
Correct Answer: B
Solution :
Key Idea: Potential energy stored in spring is responsible/or kinetic energy of ball. When spring gun is used work is done against the internal restoring force in the spring, this is stored as potential energy in the spring. \[\therefore \] \[\frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{u}^{2}}\] ?. (i) When angle of projection is \[{{45}^{o}}\], maximum horizontal range is obtained. \[{{R}_{\max }}=\frac{{{u}^{2}}\sin 2\theta }{g}\] For \[\theta ={{45}^{o}},\,2\theta ={{90}^{o}};\,\sin {{90}^{o}}=1\] \[\therefore \] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\] ?. (ii) From Eqs. (i) and (ii), we get \[{{R}_{\max }}=\frac{1}{2}m{{u}^{2}}\left( \frac{2}{mg} \right)\] \[{{R}_{\max }}=\frac{1}{2}k{{x}^{2}}\left( \frac{2}{mg} \right)\] \[{{R}_{\max }}=\frac{k{{x}^{2}}}{ng}\] Given, \[m=15\,g=15\times {{10}^{-3}}kg,\,g=10\,\,m/{{s}^{2}}\], \[k=600\,\,N/m,\,x=5\,cm=5\times {{10}^{-2}}m\] \[\therefore \] \[{{R}_{\max }}=\frac{600\times {{(5\times {{10}^{-2}})}^{2}}}{15\times {{10}^{-3}}\times 10}\] \[{{R}_{\max }}=10\,m\]
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