A wooden block of weight 40 N rests on a rough horizontal plane having a coefficient of friction 0.3. The block is struck by a bullet travelling horizontally with a velocity of 810 m/s and weighing 0.5 N. What will be the distance travelled by the block from its initial position if the bullet embedded in the block after strike? (assume, g = 10 m/s²)

Explanation:

Step 1: Calculate the initial momentum of the bullet

The initial momentum of the bullet can be calculated using the formula:

Momentum (p) = Mass (m) × Velocity (v)

The mass of the bullet (m_bullet) is:

m_bullet = W_bullet / g = 0.5 N / 10 m/s² = 0.05 kg

Given the velocity of the bullet (v_bullet) = 810 m/s, the initial momentum of the bullet is:

p_bullet = m_bullet × v_bullet = 0.05 kg × 810 m/s = 40.5 kg·m/s

Step 2: Calculate the combined mass of the block and bullet system

• Weight of the block (W_block) = 40 N

The mass of the block (m_block) is:

m_block = W_block / g = 40 N / 10 m/s² = 4 kg

The combined mass of the block and bullet system (m_total) is:

m_total = m_block + m_bullet = 4 kg + 0.05 kg = 4.05 kg

Step 3: Calculate the velocity of the block and bullet system after the collision

Using the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Total initial momentum = Total final momentum

p_bullet = m_total × v_final

Solving for v_final (the final velocity of the block and bullet system):

v_final = p_bullet / m_total = 40.5 kg·m/s / 4.05 kg = 10 m/s

When the block is moving on the rough horizontal plane, it is subjected to frictional force. The frictional force (f_friction) can be calculated using the formula:

f_friction = μ × N

where μ is the coefficient of friction and N is the normal force, Coefficient of friction (μ) = 0.3,

Normal force (N) = W_block + W_bullet = 40 N + 0.5 N = 40.5 N

The frictional force is:

f_friction = 0.3 × 40.5 N = 12.15 N

Using Newton's second law, we can calculate the deceleration (a) due to friction:

a = f_friction / m_total = 12.15 N / 4.05 kg = 3 m/s²

The block will decelerate to a stop due to this frictional force. We can use the kinematic equation to find the distance (d) travelled by the block:

v_final² = v_initial² - 2ad

Here, v_final = 0 (since the block comes to rest), v_initial = 10 m/s, and a = 3 m/s².

0 = (10 m/s)² - 2 × 3 m/s² × d

Solving for d:

100 = 6d

d = 100 / 6 = 16.67 m

Thus, the distance travelled by the block from its initial position is 16.67 meters.