Question

A bullet having a mass of 0.005 kg is movingwith a speed of 100m/s.It penetratesinto a bag of sand and is brought rest after moving 25cm into a bag .Find the declerating force on the bullet .Slso calculate the time in which it is brought to rest​

Answer 1

Explanation:

Question 19160It is given, that S=25cm=0.25m=v0ts−ats22, where tsdenotes the time for which the object moved until stop. Also, for velocity, v=v0−at, and for stop time 0=v0−ats⇒ts=v0a. Plugging this formula into formula for S, obtain: ts=2Sv0=0.005s. Hence, the acceleration is∣a∣=v0ts=100m/s0.005s=20000m/s2(actually this acceleration is negative, but one needs the absolutevalue only). Hence, ∣F∣=m∣a∣=0.005kg⋅20000m/s=100N