A bullet of mass 10 g traveling horizontally with a velocity of 150 m/s 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.
First, let's calculate the average acceleration which the bullet experiences as it travels through the block. By definition, the average acceleration is
`vec a = (vecv_f - vecv_i)/(Delta t)` , where `vec v_f` is the final velocity of the bullet (zero in this case, since the bullet comes to rest), `vec v_i` is the initial velocity (given), and `Delta t` is the time interval, also given.
Since the final velocity is zero and the initial velocity is horizontal, the average acceleration is also horizontal and directed opposite the initial velocity of the bullet. Its magnitude will be:
`a = v_i/(Delta t)`
`a = 150/0.03 = 5000 m/s^2`
The average force acting on the bullet from the block, according to the second Newton's Law, is:
`F = ma = 0.01 kg * 5000 m/s^2 = 50 N` .
The force exerted by the block on the bullet is 50 N.
To find the distance of penetration of the bullet into the block, let's use an equation of motion:
`d = v_i*Delta t - 1/2a(Delta t)^2` (Recall that acceleration comes in with the negative sign because it is directed opposite the initial velocity).
`d = 150*0.03 - 1/2*5000*(0.03)^2 = 2.25 m`
The distance of penetration of the bullet into the block is 2.25 meters.