A bullet of 10 g moving with a speed of 100 m/s penetrates asandbag and comes to rest in 1/10 th second Finda) The distance through which the bullet penetratesb) The retarding force experienced

a) To find the distance through which the bullet penetrates, we can use the formula for the distance travelled under uniform deceleration:

d = ut + 0.5at^2

where: u = initial velocity = 100 m/s t = time = 1/10 s a = acceleration (which is negative as it's deceleration)

First, we need to find the deceleration (a). We can use the formula:

v = u + at

where: v = final velocity = 0 (as the bullet comes to rest)

Rearranging the formula to find a gives:

a = (v - u) / t a = (0 - 100) / (1/10) a = -1000 m/s^2

Now we can find the distance (d):

d = ut + 0.5at^2 d = 100*(1/10) + 0.5*(-1000)*(1/10)^2 d = 10 - 5 d = 5 m

So, the bullet penetrates 5 meters into the sandbag.

b) To find the retarding force experienced, we can use Newton's second law:

F = ma

where: m = mass = 10 g = 0.01 kg (converted from grams to kilograms)

So:

F = 0.01 * -1000 F = -10 N

The negative sign indicates that the force is acting in the opposite direction to the motion of the bullet. So, the bullet experiences a retarding force of 10 N.