Question

A $4\;g$ bullet is fired horizontally with a speed of $300\;m{s^{ - 1}}$ into $0.8\;kg$ block of wood, which is at rest on a table. If the coefficient of friction between the block and the table is $0.3$ , how far will the block slide approximately?
A. $0.20\;m$
B. $0.375\;m$
C. $0.565\;m$
D. $0.750\;m$

Answer 1

Hint: A bullet moving with a kinetic energy hits the block and the block tends to move with some energy. Hence, by using the conservation of momentum, the velocity of the block will be calculated. Then, the block slides in a table which has some coefficient of friction between them. Thus, using Newton's law of motion, the distance travelled by the block on the table will be calculated.

Useful formula:
Deceleration due to friction is given by,
$a = - \mu g$
Where, $a$ is the deceleration of the object, $\mu$ is the coefficient of friction and $g$ is acceleration due to gravity.

Newton’s law of motion is given by,
${v^2} = {u^2} + 2as$
Where, $v$ is the final velocity of the object, $u$ is the initial velocity of the object, $a$ is the acceleration of the object and $s$ is the distance travelled by the object.

Given data:
The mass of the bullet, ${m_1} = 4\;g$
The mass of the block, ${m_2} = 0.8\;kg = 800\;g$
The speed of the bullet, ${v_1} = 300\;m{s^{ - 1}}$
The speed of the block after bullet hitting it is ${v_2}$
The coefficient of friction between the block and the table, $\mu = 0.3$
The acceleration due to gravity, $g = 9.8\;m{s^{ - 2}}$

Complete step by step solution:
By applying the conservation of momentum in bullet hitting the block, we get
${m_1}{v_1} = {m_2}{v_2}$
By substituting the given values, we get
$4 \times 300 = 800 \times {v_2} \\\ {v_2} = \dfrac{{4 \times 300}}{{800}} \\\ {v_2} = 1.5\;m{s^{ - 1}} \\\$

The block tends to move on the table with a coefficient of friction $\mu = 0.3$.
Hence, the block tends to decelerate, the deceleration is
$a = - \mu g$
Substitute the given values, we get
$a = - 0.3 \times 9.8 \\\ a = - 2.94\;m{s^{ - 2}} \\\$

By Newton’s law of motion,
${u_2}^2 = {v_2}^2 + 2as$
Since, finally the block tends to rest. Then, the final velocity of the block is zero, ${u_2} = 0$
Substituting the given values, we get
${0^2} = {\left( {1.5\;m{s^{ - 1}}} \right)^2} + \left( {2 \times \left( { - 2.94\;m{s^{ - 2}}} \right) \times s} \right) \\\ 0 = 2.25 - \left( {5.88 \times s} \right) \\\ 2.25 = 5.88 \times s \\\ s = \dfrac{{2.25}}{{5.88}} \\\ s = 0.382\;m \\\ s \simeq 0.375\;m \\\$

Hence, the option (B) is correct.

Note: The force of the bullet tends to move the block with an initial velocity above the table. The frictional force between the block and the table tends the block to the state of rest. Hence, the distance travelled by the block is completely dependent on the force of the bullet and the coefficient of friction between the block and the table.