
A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distance will it penetrate into the target before it comes to rest?
1) 30 cm
2) 20 cm
3) 10 cm
4) 5 cm
Answer
163.5k+ views
Hint: Firstly, we should know about the concept of work energy theorem. According to the work-energy theorem, also referred to as the principle of work and kinetic energy, the total work performed by all forces acting on a particle is equal to the change in that particle's kinetic energy.
Complete answer:
Let the velocity of bullet fired is v
Given that the velocity decreases to half after penetrating 30 cm, i.e $v= \dfrac {v}{2}$
Suppose s be the distance travelled by the bullet through 30 cm and F be the retarding force offered by the target.
According to work energy theorem, we can write the change in kinematic energy as:
$\dfrac{1}{2}m{{(\dfrac{v}{2})}^{2}}-\dfrac{1}{2}m{{v}^{2}}=-F.s$
$i.e.\dfrac{1}{2}m{{v}^{2}}=\dfrac{4}{3}F.s............................(1)$
Let’s be the distance travelled by the bullet inside the target when the velocity of the bullet becomes zero i.e. v becomes 0.
According to work energy theorem,
$\dfrac{1}{2}m\times {{0}^{2}}-\dfrac{1}{2}m{{v}^{2}}=-F.{s}'$
$i.e.\dfrac{1}{2}m{{v}^{2}}=F.{s}'.......................(2)$
From (1) and (2) equation, we get:
$\dfrac{4}{3}F.s=F.{s}'$
${s}'=\dfrac{4}{3}\times s$
Given, s = 30 cm
$\Rightarrow {s}'=\dfrac{4}{3}\times 30$
${s}'=40cm$
So the additional thickness the bullet will penetrate before coming to rest will be:
$\Rightarrow {s}'-s$
$\Rightarrow 40-30=10cm$
Hence, option (3) is correct.
Note: We may use the Work-Energy theorem to determine a particle's velocity given a known force at any place even if its full range of applications won't become clear until we examine energy conservation. This skill is advantageous because it connects our derived concept of work to elementary kinematics.
Complete answer:
Let the velocity of bullet fired is v
Given that the velocity decreases to half after penetrating 30 cm, i.e $v= \dfrac {v}{2}$
Suppose s be the distance travelled by the bullet through 30 cm and F be the retarding force offered by the target.
According to work energy theorem, we can write the change in kinematic energy as:
$\dfrac{1}{2}m{{(\dfrac{v}{2})}^{2}}-\dfrac{1}{2}m{{v}^{2}}=-F.s$
$i.e.\dfrac{1}{2}m{{v}^{2}}=\dfrac{4}{3}F.s............................(1)$
Let’s be the distance travelled by the bullet inside the target when the velocity of the bullet becomes zero i.e. v becomes 0.
According to work energy theorem,
$\dfrac{1}{2}m\times {{0}^{2}}-\dfrac{1}{2}m{{v}^{2}}=-F.{s}'$
$i.e.\dfrac{1}{2}m{{v}^{2}}=F.{s}'.......................(2)$
From (1) and (2) equation, we get:
$\dfrac{4}{3}F.s=F.{s}'$
${s}'=\dfrac{4}{3}\times s$
Given, s = 30 cm
$\Rightarrow {s}'=\dfrac{4}{3}\times 30$
${s}'=40cm$
So the additional thickness the bullet will penetrate before coming to rest will be:
$\Rightarrow {s}'-s$
$\Rightarrow 40-30=10cm$
Hence, option (3) is correct.
Note: We may use the Work-Energy theorem to determine a particle's velocity given a known force at any place even if its full range of applications won't become clear until we examine energy conservation. This skill is advantageous because it connects our derived concept of work to elementary kinematics.
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