Question

A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find the deceleration.

Answer 1

A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. The deceleration is $-1225000 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

Step 1:

From the given statements, we know

Initial Velocity (u) = 350 m/s

Final Velocity (v)  = 0 m/s

Distance traveled (s) = 5 cm = 0.05 m

Step 2:

To find retardation (a), We Substitute the values in Newton’s 3rd Equation of Motion

$v^{2}=u^{2}+2 a s$

Substituting the values in the above Equation

    $0 = (350)2 + 2\times a\times0.05$

    $2\times a\times 0.05 = -(350\times 350)$

    $a = -350\times 350\times 10$

       $=-1225000 \mathrm{m} / \mathrm{s}^{2}$