A) \[\frac{{{n}^{2}}}{2n-1}\]
B) \[\frac{2{{n}^{2}}}{n-1}\]
C) infinite
D) n
Correct Answer: A
Solution :
Let u be the initial velocity of the bullet of mass m. After passing through a plank of width x, its velocity decreases to v. \[\therefore \]\[u-v=\frac{4}{n}\]or, \[v=u-\frac{4}{n}=\frac{u(n-1)}{n}\] If F be the retarding force applied by each plank, then using work ? energy theorem, \[Fx=\frac{1}{2}m{{u}^{2}}-\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{u}^{2}}-\frac{1}{2}m{{u}^{2}}\frac{{{\left( n-1 \right)}^{2}}}{{{n}^{2}}}\] \[=\frac{1}{2}m{{u}^{2}}\left[ \frac{1-{{\left( n-1 \right)}^{2}}}{{{n}^{2}}} \right]\] \[Fx=\frac{1}{2}m{{u}^{2}}\left( \frac{2n-1}{{{n}^{2}}} \right)\] Let P be the number of planks required to stop the bullet. Total distance travelled by the bullet before coming to rest = Px Using work-energy theorem again, \[F(Px)=\frac{1}{2}m{{u}^{2}}-0\] or, \[P(Fx)=P\left[ \frac{1}{2}m{{u}^{2}}\frac{(2n-1)}{{{n}^{2}}} \right]=\frac{1}{2}m{{u}^{2}}\] \[\therefore \]\[P=\frac{{{n}^{2}}}{2n-1}\]
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