Answer
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Hint: This problem is related to the motion of the object. Thus, we have to find the Kinetic energy of the planks. So, find the kinetic energy lost due to one plank and then using the kinetic energy obtained find the total energy lost due to N planks. Now, substitute the value of energy lost due to one plank in the expression for total energy lost. Thus, calculate N i.e. total number of planks required to stop the bullet.
Complete answer:
Let the initial speed of bullet be v
The bullet loses $({{\dfrac {1}{n}}})^{th}$ of its velocity passing through one plank
$\Rightarrow new \quad velocity= v- \dfrac {v}{n}$
As the velocity decreases, energy gets lost.
Therefore, Kinetic energy lost is given by,
$\Delta E= \dfrac {1}{2} m{v}^{2}$
Substituting values in above equation we get,
$\Delta E= \dfrac {1}{2} m{v}^{2} - \dfrac {1}{2}m({ v- \dfrac {v}{n}})^{2}$
$\Rightarrow \Delta E= \dfrac {1}{2} m{v}^{2}(1- ({\dfrac {n-1}{n}})^{2})$
Now, let the number of planks after which velocity becomes 0 be N.
The record, total energy lost is given by,
${E}_{T} = N × E$
$\Rightarrow \dfrac {1}{2} m{v}^{2}= N × E$
Substituting value of E in above equation we get,
$ \dfrac {1}{2} m{v}^{2}= N × \dfrac {1}{2} m{v}^{2}(1- ({ \dfrac{n-1}{n}})^{2})$
Rearranging above equation we get,
$N= (1- ({ \dfrac{n-1}{n})}^{2})$
$\Rightarrow N= \dfrac {{n}^{2}}{2n-1}$
Thus, $\dfrac {{n}^{2}}{2n-1}$ planks will be required to stop the bullet.
Hence, the correct answer is option A i.e. $\dfrac {{n}^{2}}{2n-1}$.
Note: In this problem, as the velocity decreases consequently acceleration decreases too. Thus, we can solve this problem using Kinematics equation as well i.e. third equation of motion i.e. $ { v }_{ f }^{ 2 }={ v }_{ i }^{ 2 }+2as$
Where, ${v}_{i}$ is the initial velocity
${v}_{f}$ is the final velocity
a is the constant acceleration
s is the displacement
As the number of planks increases, velocity of the bullet decreases. Hence, the energy also decreases with increase in the number of planks.
Complete answer:
Let the initial speed of bullet be v
The bullet loses $({{\dfrac {1}{n}}})^{th}$ of its velocity passing through one plank
$\Rightarrow new \quad velocity= v- \dfrac {v}{n}$
As the velocity decreases, energy gets lost.
Therefore, Kinetic energy lost is given by,
$\Delta E= \dfrac {1}{2} m{v}^{2}$
Substituting values in above equation we get,
$\Delta E= \dfrac {1}{2} m{v}^{2} - \dfrac {1}{2}m({ v- \dfrac {v}{n}})^{2}$
$\Rightarrow \Delta E= \dfrac {1}{2} m{v}^{2}(1- ({\dfrac {n-1}{n}})^{2})$
Now, let the number of planks after which velocity becomes 0 be N.
The record, total energy lost is given by,
${E}_{T} = N × E$
$\Rightarrow \dfrac {1}{2} m{v}^{2}= N × E$
Substituting value of E in above equation we get,
$ \dfrac {1}{2} m{v}^{2}= N × \dfrac {1}{2} m{v}^{2}(1- ({ \dfrac{n-1}{n}})^{2})$
Rearranging above equation we get,
$N= (1- ({ \dfrac{n-1}{n})}^{2})$
$\Rightarrow N= \dfrac {{n}^{2}}{2n-1}$
Thus, $\dfrac {{n}^{2}}{2n-1}$ planks will be required to stop the bullet.
Hence, the correct answer is option A i.e. $\dfrac {{n}^{2}}{2n-1}$.
Note: In this problem, as the velocity decreases consequently acceleration decreases too. Thus, we can solve this problem using Kinematics equation as well i.e. third equation of motion i.e. $ { v }_{ f }^{ 2 }={ v }_{ i }^{ 2 }+2as$
Where, ${v}_{i}$ is the initial velocity
${v}_{f}$ is the final velocity
a is the constant acceleration
s is the displacement
As the number of planks increases, velocity of the bullet decreases. Hence, the energy also decreases with increase in the number of planks.
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