Question

A gun is fired at a 3cm thick solid wooden door.the bullet of mass 7g travels through the door and has its speed reduced from 450m/s to 175m/s .
assuming uniform resistance.what is the force of the wood on the bullet

Answer 1

Answer:

Let, the initial velocity of the bullet is u = 450m/sec

       the final velocity of the bullet is v = 175m/sec

       The thickness of the solid wooden door is s = 3cm = 0.03m

       The mass of the bullet is m = 7g = 0.007kg

Using the equation, v^2 - u^2 = 2as

                                 a = v^2 - u^2 / 2s

                                 a = (175)^2 - (450)^2 / 2*0.03

                                 a = 30625 - 202500 / 0.06

                                 a = -171875 / 0.06

                                 a = -2864583.333

since acceleration is never negative so we will consider the value positive.

Now, from the equation F = ma we have,

                                       F = 0.007 * 2864583.333

                                       F = 20052. 083

The resistive force on wood is 20052.083N.

Explanation: