Solveeit Logo
Login

Question

Question: A bullet of mass m losses \({{\left( \dfrac{1}{n} \right)}^{\text{th}}}\) of its velocity passing th...

A bullet of mass m losses (1n)th{{\left( \dfrac{1}{n} \right)}^{\text{th}}} of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be:
A. n22n−1 B. 2n2n−1 C. infinite D. n \begin{aligned} & \text{A}\text{. }\dfrac{{{n}^{2}}}{2n-1} \\\ & \text{B}\text{. }\dfrac{2{{n}^{2}}}{n-1} \\\ & \text{C}\text{. infinite} \\\ & \text{D}\text{. n} \\\ \end{aligned}

Explanation

Solution

From the velocity loss of the bullet when passing through the plank, obtain the expression for the width of the plank through which the bullet is passed using the laws of motion. Consider the N number plank used to stop the bullet. Then use the required laws of motion and put the required values to find the solution to this question.

Complete answer:
The mass of the bullet is m.
The bullet losses (1n)th{{\left( \dfrac{1}{n} \right)}^{\text{th}}} of its velocity passing through one plank.
Let the initial velocity of the bullet is u and the velocity of the bullet after passing through the plank will be, v=u−unv=u-\dfrac{u}{n}
Let the retardation of the bullet is a and the width of the plank is s.
Then we can write,
v2=u2+2as ⇒(u−un)2=u2+2as ⇒u2−2u2n+u2n2=u2+2as ⇒2as=u2n2−2u2n ⇒2as=u2(1n2−2n) ⇒s=u2(1n2−2n)2a \begin{aligned} & {{v}^{2}}={{u}^{2}}+2as \\\ & \Rightarrow {{\left( u-\dfrac{u}{n} \right)}^{2}}={{u}^{2}}+2as \\\ & \Rightarrow {{u}^{2}}-2\dfrac{{{u}^{2}}}{n}+\dfrac{{{u}^{2}}}{{{n}^{2}}}={{u}^{2}}+2as \\\ & \Rightarrow 2as=\dfrac{{{u}^{2}}}{{{n}^{2}}}-2\dfrac{{{u}^{2}}}{n} \\\ & \Rightarrow 2as={{u}^{2}}\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right) \\\ & \Rightarrow s=\dfrac{{{u}^{2}}\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right)}{2a} \\\ \end{aligned}
Let the number of planks is N.
If the retardation remains constant, then we can write that after passing through N planks the final velocity of the bullet will be zero.
Width of one plank is s.
So, the width of N plank is =Ns=Ns
Now, we can write,
v2=u2+2aNs ⇒0=u2+2aNu2(1n2−2n)2a ⇒0=u2+Nu2(1n2−2n) ⇒0=1+N(1n2−2n) ⇒1=−N(1n2−2n) ⇒1N=2n−1n2 ⇒1N=2n−1n2 ⇒N=n22n−1 \begin{aligned} & {{v}^{2}}={{u}^{2}}+2aNs \\\ & \Rightarrow 0={{u}^{2}}+2aN\dfrac{{{u}^{2}}\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right)}{2a} \\\ & \Rightarrow 0={{u}^{2}}+N{{u}^{2}}\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right) \\\ & \Rightarrow 0=1+N\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right) \\\ & \Rightarrow 1=-N\left( \dfrac{1}{{{n}^{2}}}-\dfrac{2}{n} \right) \\\ & \Rightarrow \dfrac{1}{N}=\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}} \\\ & \Rightarrow \dfrac{1}{N}=\dfrac{2n-1}{{{n}^{2}}} \\\ & \Rightarrow N=\dfrac{{{n}^{2}}}{2n-1} \\\ \end{aligned}
So, the total number of planks needed to stop the bullet will be n22n−1\dfrac{{{n}^{2}}}{2n-1}.

The correct answer is option (A).

Note:
The laws of motion needed to find the solution to this question gives the relation between the final velocity, initial velocity, acceleration and the distance covered by a particle. Here, the velocity of the particle is decreasing with time and we have deceleration in place of acceleration. We can use it as a negative acceleration or we can simply put the value of the deceleration.