Question

if A bullet of 10 gram strikes a sandbag at a speed of 500 m per second and gets embedded after travelling 5 cm find the resistive force exerted by the send on the bullet and time taken by the bullet to come at 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 a=v-u/t

-10000000=0-1000/t

-10000000=-1000/t

-10000=t

or

t=-10⁴second

Answer 2

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 a=v-u/t

-10000000=0-1000/t

-10000000=-1000/t

-10000=t

or

t=-10⁴seconds

HOPE THIS WILL HELP YOU..........................................................