Question

: A bullet of mass $5\;{\rm{g}}$ travelling with a velocity of .${\rm{18}}\;{\rm{km}} \cdot {\rm{h}}{{\rm{r}}^{ - 1}}$. penetrates a wooden block and comes to rest in $0.01\;{\rm{s}}$. Find
(a) Find the distance it penetrates in the wooden block.
(b) The retarding force of the block.

Answer 1

To obtain the desired solutions, we can use Newton's equations of motion. We can obtain the value of retarding force using Newton’s second law of motion.

Complete step by step answer:
(a) It is given that a bullet of mass $5\;{\rm{g}}$ moving with a velocity of ${\rm{18}}\;{\rm{km}} \cdot {\rm{h}}{{\rm{r}}^{ - 1}}$ penetrates a wooden block and comes to rest in $0.01\;{\rm{s}}$. Let $m$ be the mass of the bullet and $v$ is the initial velocity with which it was travelling. Let $t$ be the time taken by the bullet to pass through the block after it penetrated the block. Hence, we can write
$m = 5g\\\ \Rightarrow v = {\rm{18}}\;{\rm{km}} \cdot {\rm{h}}{{\rm{r}}^{ - 1}}\\\ \Rightarrow t = 0.01\;{\rm{s}}$
We will convert the values of velocity and mass into SI units.
Hence, we can write the value of mass as
$m = 5\;{\rm{g}} \times \dfrac{{{{10}^{ - 3}}\;{\rm{kg}}\;}}{{1\;{\rm{g}}}}\\\ \Rightarrow m = 5 \times {10^{ - 3}}\;{\rm{kg}}$
The value of initial velocity in SI units is written as

$$v = {\rm{18}}\;{\rm{km}} \cdot {\rm{h}}{{\rm{r}}^{ - 1}} \times \dfrac{{\dfrac{5}{{18}}\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}}}{{1\;{\rm{km}} \cdot {\rm{h}}{{\rm{r}}^{ - 1}}\;}}\\\ \Rightarrow v = 5\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$$

The velocity of the bullet which penetrated the block becomes zero after the bullet comes to rest. Hence, we can say that the final velocity of the bullet is zero. Let $v$ be the final velocity of the bullet. Then,
$v = 0$
Since the values for the initial velocity $u$, the final velocity $v$ and the time taken by the bullet to move through the block before coming to rest are known, we can use the Newton’s equation of motion,$v = u + at$, to find the acceleration of the bullet. Now, substituting the values of$u$, $v$ and $t$ in the equation $v = u + at$, we get

$$0 = 5\;\; + a \times 0.01\;\\\ \Rightarrow a \times 0.01\; = - 5\;\\\ \Rightarrow a = \dfrac{{ - 5\;}}{{0.01\;}}\\\ \Rightarrow a = - 500\;{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$$

Since the bullet comes to rest, the bullet undergoes deceleration, So, we obtained a negative value for the acceleration.

Since we now know the value of acceleration, we can obtain the distance penetrated by the bullet in the block using the Newton’s law equation written as,
${v^2} - {u^2} = 2aS$
Here $S$ is the distance penetrated by the bullet through the block.

We can substitute the values of $u$, $v$ and $a$ in the above equation to obtain the value of the distance$S$. Since $u = 5\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$, $v = 0\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ and $a = - 500\;{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$, we write
${0^2} - {5^2} = 2 \times \left( { - 500} \right) \times S\\\ \Rightarrow - 25 = - 1000S\\\ \Rightarrow\dfrac{{25}}{{1000}} = S\\\ \Rightarrow 0.025\;{\rm{m}} = S$
Hence, the bullet penetrated $0.025\;{\rm{m}}$ distance in the block.

The answer for subpart (a) is $0.025\;{\rm{m}}$.

(b) From Newton’s second law of motion, the retarding force is the product of mass and acceleration. Hence, we can write
$F = ma$
Here $F$ is the retarding force.

Let’s substitute $5 \times {10^{ - 3}}\;{\rm{kg}}$ for $m$ and $- 500\;{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for $a$ in $F = ma$, we get

$$F = 5 \times {10^{ - 3}}\;{\rm{kg}} \times \left( { - 500\;{\rm{m}} \cdot {{\rm{s}}^{ - 2}}} \right)\\\ \therefore F= - 2.5\;{\rm{N}}$$

Here the negative sign implies that the force is opposite to the direction of motion of the bullet.Hence, we obtained the magnitude of the retarding force as $2.5\;{\rm{N}}$.

Thus, the answer for subpart (b) is $2.5\;{\rm{N}}$.

Note: Please note that if an object is at rest at its final position, then we should take the final velocity of the object as zero. Also, do not omit the negative sign obtained in the deceleration as it will affect the calculation.