Question

A bullet of mass \[20\,{\text{g}}\] has an initial speed of \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\], just before it starts penetrating a mud wall of thickness \[20\,{\text{cm}}\]. If the wall offers a mean resistance of \[2.5 \times {10^{ - 2}}\,{\text{N}}\], the speed of the bullet after emerging from the other side of the wall is close to:
A. \[0.4\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
B. \[0.1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
C. \[0.3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
D. \[0.7\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]

Answer 1

Hint:Use the formula of force on an object. This formula gives the relation between the force on the object, mass of the object and acceleration of the object. Also, use the kinematic equation relating the final velocity, initial velocity, acceleration and displacement of the object. Determine the acceleration of the bullet and substitute it in the kinematic equation to determine the final velocity of the bullet.

Formulae used:
The force \[F\] acting on an object is given by
\[F = ma\] …… (1)
Here, \[m\] is the mass of the object and \[a\] is the acceleration of the object.
The third kinematic equation for the final velocity of the object is given by
\[{v^2} = {u^2} + 2as\] …… (2)
Here, \[v\] is the final velocity, \[u\] is the initial velocity, \[a\] is the acceleration and \[s\] is the displacement of the object.

Complete step by step answer:
We have given that the mass and initial velocity of the bullet are \[20\,{\text{g}}\] and \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] respectively.
\[m = 20\,{\text{g}}\]
\[u = 1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
The resistance force offered by the mud wall to the bullet is \[2.5 \times {10^{ - 2}}\,{\text{N}}\].
\[F = 2.5 \times {10^{ - 2}}\,{\text{N}}\]
The thickness of the mud wall is \[20\,{\text{cm}}\]. The bullet penetrates this mud wall. Hence, the displacement of the bullet is \[20\,{\text{cm}}\].
\[s = 20\,{\text{cm}}\]
We should first calculate the acceleration of the bullet when it enters the mud wall.
Rearrange the equation (1) for the acceleration \[a\] of the bullet.
\[a = \dfrac{F}{m}\]
Substitute \[ - 2.5 \times {10^{ - 2}}\,{\text{N}}\] for \[F\] and \[20\,{\text{g}}\] for \[m\] in the above equation.
\[a = \dfrac{{ - 2.5 \times {{10}^{ - 2}}\,{\text{N}}}}{{20\,{\text{g}}}}\]
\[ \Rightarrow a = - \dfrac{{2.5 \times {{10}^{ - 2}}\,{\text{N}}}}{{20 \times {{10}^{ - 3}}\,{\text{kg}}}}\]
\[ \Rightarrow a = - 1.25\,{\text{m/}}{{\text{s}}^2}\]
Hence, the acceleration of the bullet through the mud wall is \[ -
1.25\,{\text{m/}}{{\text{s}}^2}\].
We can determine the final velocity of the bullet using equation (2).
Substitute \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[u\], \[ - 1.25\,{\text{m/}}{{\text{s}}^2}\] for \[a\] and \[20\,{\text{cm}}\] for \[s\] in equation (2).
\[{v^2} = {\left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}u} \right)^2} + 2\left( { - 1.25\,{\text{m/}}{{\text{s}}^2}} \right)\left( {20\,{\text{cm}}} \right)\]
\[ \Rightarrow {v^2} = {\left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}u} \right)^2} + 2\left( { - 1.25\,{\text{m/}}{{\text{s}}^2}} \right)\left[ {\left( {20\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]\]
\[ \Rightarrow {v^2} = 1 - 0.5\]
\[ \Rightarrow {v^2} = 0.5\]
Take square root on both sides of the above equation.
\[ \Rightarrow v = \sqrt {0.5} \]
\[ \therefore v = 0.7\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
Therefore, the velocity of the bullet after emerging out from the other end of the wall is \[0.7\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\].

Hence, the correct option is D.

Note: One may think why the resistance force offered by the wall is taken negative. The mud wall opposes the motion of the bullet and exerted the resistance force in a direction opposite to that of the motion of the bullet. Hence, the sign of the resistance force is taken negative. The velocity of the bullet decreases after passing through the mud wall. As a result, its acceleration is negative.