Question

A bullet of mass 10 g traveling horizontally with a velocity of 150 m s^–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Given: 

Mass of bullet, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \)
Initial velocity, \( u = 150 \, \text{m/s} \)
Final velocity, \( v = 0 \, \text{m/s} \)
Time to stop, \( t = 0.03 \, \text{s} \)

Step 1: Calculate acceleration

Using the formula: \( a = \frac{v - u}{t} \)
\[ a = \frac{0 - 150}{0.03} = -5000 \, \text{m/s²} \] (negative sign indicates deceleration)

Step 2: Calculate distance of penetration

Using the equation of motion: \( s = ut + \frac{1}{2} a t^2 \)
\[ s = 150 \times 0.03 + \frac{1}{2} \times (-5000) \times (0.03)^2 \]
\[ s = 4.5 - 2.25 = 2.25 \, \text{m} \]

Step 3: Calculate force exerted by the block

Using Newton’s second law: \( F = m a \)
\[ F = 0.01 \times 5000 = 50 \, \text{N} \]

Answer:

Distance of penetration = 2.25 m
Magnitude of force = 50 N