Question

A bullet moving with a speed of $150{\text{m}}{{\text{s}}^{ - 1}}$ strikes a wooden plank. After passing through the plank, its speed becomes $125{\text{m}}{{\text{s}}^{ - 1}}$ . Another bullet of the same mass and size strikes the plank with a speed of $90{\text{m}}{{\text{s}}^{ - 1}}$ . Find its speed after passing through the plank.
A) $25{\text{m}}{{\text{s}}^{ - 1}}$
B) $35{\text{m}}{{\text{s}}^{ - 1}}$
C) $50{\text{m}}{{\text{s}}^{ - 1}}$
D) ${\text{70m}}{{\text{s}}^{ - 1}}$

Answer 1

Hint:As the bullet passes through the wooden plank, its velocity decreases or kinetic energy decreases i.e., energy is lost. This loss in energy is due to the work done against the wooden plank. The work-energy theorem states that the work done by the net force will be equal to the change in kinetic energy of the body.

Formula used:
-The change in kinetic energy of a body is given by, $\Delta K = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)$ where $m$ is the mass of the body, $v$ is its final velocity and $u$ is its initial velocity.

Complete step by step answer.
Step 1: Sketch a rough figure of the problem and list the parameters known from the question.

Two bullets of the same mass and size pass through a wooden plank one after the other.
The initial velocity of the first bullet is ${u_1} = 150{\text{m}}{{\text{s}}^{ - 1}}$ and its final velocity is ${v_1} = 125{\text{m}}{{\text{s}}^{ - 1}}$ .
The initial velocity of the second bullet is ${u_1} = 90{\text{m}}{{\text{s}}^{ - 1}}$ but its final velocity ${v_1}$ is unknown.
Let the mass of the two bullets be represented by $m$ .
Step 2: Find the change in kinetic energies of the first bullet and the second bullet.
The change in kinetic energy of a body is given by, $\Delta K = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)$ -------- (1) where $m$ is the mass of the body, $v$ is its final velocity and $u$ is its initial velocity.
First bullet
Using equation (1) the change in kinetic energy of the first bullet can be expressed as $\Delta {K_1} = \dfrac{1}{2}m\left( {{v_1}^2 - {u_1}^2} \right)$
where $m$ is the mass of the bullet, ${v_1}$ is its final velocity and ${u_1}$ is its initial velocity.
Substituting for ${u_1} = 150{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_1} = 125{\text{m}}{{\text{s}}^{ - 1}}$ in the above expression we get, $\Delta {K_1} = \dfrac{1}{2}m\left( {{{125}^2} - {{150}^2}} \right)$ ------- (2)
Second bullet
Similarly, using equation (1) the change in kinetic energy of the second bullet can be expressed as $\Delta {K_2} = \dfrac{1}{2}m\left( {{v_2}^2 - {u_2}^2} \right)$
where $m$ is the mass of the bullet, ${v_2}$ is its final velocity and ${u_2}$ is its initial velocity.
Substituting for ${u_2} = 90{\text{m}}{{\text{s}}^{ - 1}}$ in the above expression we get, $\Delta {K_2} = \dfrac{1}{2}m\left( {{v_2}^2 - {{90}^2}} \right)$ ------- (3)
Equations (2) and (3) respectively represent the change in kinetic energy of the first and second bullet.
Step 3: Find the final velocity of the second bullet based on the work-energy theorem.
The bullet on passing through the wooden plank suffers a decrease in its kinetic energy. This loss in kinetic energy contributes to the work done against the plank.
According to the work-energy theorem, the work done against the wooden plank must be equal to the change in the kinetic energy of the bullet.
i.e., ${W_p} = \Delta K$
For the first bullet, the work done against the plank will be ${W_{p1}} = \Delta {K_1} = \dfrac{1}{2}m\left( {{{125}^2} - {{150}^2}} \right)$
For the second bullet, the work done against the plank will be ${W_{p2}} = \Delta {K_2} = \dfrac{1}{2}m\left( {{v_2}^2 - {{90}^2}} \right)$
The work done against the plank will be the same for both bullets i.e., ${W_{p1}} = {W_{p2}}$ .
Thus we can equate equations (2) and (3).
We have, $\dfrac{1}{2}m\left( {{{125}^2} - {{150}^2}} \right) = \dfrac{1}{2}m\left( {{v_2}^2 - {{90}^2}} \right)$
Cancelling out the similar terms on either side of the above equation we get, $\left( {{{125}^2} - {{150}^2}} \right) = \left( {{v_2}^2 - {{90}^2}} \right)$
This can be rearranged to get, ${v_2}^2 = {125^2} - {150^2} + {90^2}$
Now, taking the squaring root on both sides we get, ${v_2} = \sqrt {{{125}^2} - {{150}^2} + {{90}^2}} = 35{\text{m}}{{\text{s}}^{ - 1}}$
So, the velocity of the second bullet on passing through the wooden plank is ${v_2} = 35{\text{m}}{{\text{s}}^{ - 1}}$ .

Thus the correct option is B.

Note: Alternate method
The velocity of the second bullet can also be found out using Newton’s third equation of motion given by, ${v^2} - {u^2} = 2as$ ----- (a) where $v$ is the final velocity of the body, $u$ is its initial velocity, $a$ is its acceleration and $s$ is the distance covered.
Here, the distance covered refers to the thickness of the plank and the deceleration is the same for both bullets.
For the first bullet, equation (a) can be expressed as ${125^2} - {150^2} = 2as = - 6875$ ------- (b) since the final velocity of the first bullet is ${v_1} = 125{\text{m}}{{\text{s}}^{ - 1}}$ and its initial velocity is ${u_1} = 150{\text{m}}{{\text{s}}^{ - 1}}$ .
The negative sign indicates that the bullet decelerates.
For the second bullet, equation (a) can be expressed as ${v_2}^2 - {90^2} = 2as$ ------- (c) since the final velocity of the second bullet is ${v_2}$ and its initial velocity is ${u_2} = 90{\text{m}}{{\text{s}}^{ - 1}}$ .
Substituting the value for $2as = - 6875$ (from equation (b)) in equation (c) we get, ${v_2}^2 - {90^2} = - 6875$
Then we have, ${v_2}^2 = - 6875 + {90^2} = 1225$ and on taking its square root we get, ${v_2} = 35{\text{m}}{{\text{s}}^{ - 1}}$
Thus the velocity of the second bullet as it passes through the plank is ${v_2} = 35{\text{m}}{{\text{s}}^{ - 1}}$ .