Question

Using second law of motion, derive the relation between force and acceleration. A bullet of 10 gm strikes a sand bag at a speed of 1000 m/s and gets embedded after travelling 5 cm. Calculates. (i) resistive force exerted by sand on bullet (ii) the time taken by bullet to come to rest.

Answer 1

Mass of bullet should be =10gm=0.01kg

initial velocity (U)of bullet=1000m/s

final velocity (V)=0m/s(because it has stopped after striking sand)

displacement of bullet=5cm=0.05m

a)using 2as=V ²-U²

2(a)(0.05)=0²-1000²

2a(5/100)=-1000000

2a(1/20)=-1000000

a(1/10)=-1000000

a=-10000000=-10power7

acceleration of bullet is -10power7or -10000000

by using F=ma

F=0.01(10000000)

F=1/100(10000000)

F=-100000N

there fore resistive force of bullet =-100000N

b)by using t=v-u/a

t=-1000/-10^7

t=10^-4sec