Question

A bullet when fired at a target with a velocity of $100\;m{s^{ - 1}}$ penetrates one meter into it. If the bullet is fired at a similar target with a thickness $0.5\;m$. Then, at what velocity the bullet will emerge from it?
A. $50\sqrt 2 \;m{s^{ - 1}}$
B. $\dfrac{{50}}{{\sqrt 2 }}\;m{s^{ - 1}}$
C. $50\;m{s^{ - 1}}$
D. $10\;m{s^{ - 1}}$

Answer 1

Hint: In the first condition, the bullet penetrates the target to one meter. Hence, the target has some work done to stop the bullet. By law of conservation of energy, the kinetic energy of the bullet is equal to work done by the bullet. In the second case, the similar target work done with $0.5\;m$ bullet penetration. Thus, the work done by the wall is equal to the difference in the kinetic energy of the two cases. By using this relation, the velocity can be calculated.

Useful formula:
Work done, $W = F \times d$
Where, $W$ is the work done, $F$ is the force applied and $d$ is the distance travelled.

Kinetic energy, $K.E = \dfrac{1}{2}m{v^2}$
Where, $K.E$ is the kinetic energy of the object, $m$ is the mass of the object and $v$ is the velocity of the object.

Given data:
The velocity of bullet in first case, ${v_1} = 10\;m{s^{ - 1}}$
The distance travelled by bullet in first case, ${d_1} = 1\;m$
The distance travelled by bullet in second case, ${d_2} = 0.5\;m$

Complete step by step solution:
Assume that the force applied by the bullet on target is $F$
In the first case, the work done by the target to stop the bullet,
${W_1} = F \times {d_1}$
Substitute the given values, we get
$
  {W_1} = F \times 1 \\
  {W_1} = F \\
 $
The kinetic energy,
$K.{E_1} = \dfrac{1}{2}m{v_1}^2$
Where, $K.{E_1}$ is the kinetic energy of a bullet in the first case.
Substitute the given values, we get
$K.{E_1} = \dfrac{1}{2}m{\left( {100\;m{s^{ - 1}}} \right)^2}$
By taking square to the value inside the bracket, we get
$
  K.{E_1} = \dfrac{1}{2}m \times 10000 \\
  K.{E_1} = 5000m \\
 $
By law of conservation of energy,
$K.{E_1} = {W_1}$
Substitute the given values, we get
$F = 5000m\;...................................\left( 1 \right)$

In the second case, the work done by the target to stop the bullet,
${W_2} = F \times {d_2}$
Substitute the given values, we get
$
  {W_2} = 5000m \times 0.5 \\
  {W_2} = 2500 \times m \\
 $
The kinetic energy,
$K.{E_2} = \dfrac{1}{2}m{v_2}^2$
Where, $K.{E_2}$ is the kinetic energy of the bullet in the second case.
Now, the work done in second case is equal to difference in kinetic energy,
${W_2} = K.{E_1} - K.{E_2}$
Substitute the values in above equation, we get
$2500 \times m = 5000m - \dfrac{1}{2}m{v_2}^2$
Taking the term $m$ outside the bracket, it is common in RHS, we get
$2500 \times m = m\left( {5000 - \dfrac{{{v_2}^2}}{2}} \right)$
By cancelling $m$ on both sides of the equation,
$
  2500 = 5000 - \dfrac{{{v_2}^2}}{2} \\
  \dfrac{{{v_2}^2}}{2} = 5000 - 2500 \\
 $
By performing arithmetic operation on RHS, we get
$\dfrac{{{v_2}^2}}{2} = 2500$
By taking the term $2$ from denominator of RHS to the LHS, we get
$
  {v_2}^2 = 2 \times 2500 \\
  {v_2}^2 = 5000 \\
 $
Taking square root on both sides, we get
$\sqrt {{v_2}^2} = \sqrt {5000} $
By cancelling square and square root on LHS of the equation, we get
${v_2} = \sqrt {5000} $
By converting the term $5000$ into $2500 \times 2$, we get
$
  {v_2} = \sqrt {2500 \times 2} \\
  {v_2} = 50\sqrt 2 \;m{s^{ - 1}} \\
 $
Hence, the option (A) is correct.

Note: The bullet hits the target with some force creates impact on the target and the target tends to some work done to stop the bullet in certain penetration. In the second case, the bullet hits with the same force but the target thickness is $0.5\;m$, so the bullet penetrates completely and flows out with certain kinetic energy.