Question

A bullet of mass 2g moving at 400m/s strikes a wooden parcel and panel rates a distance of 5cm before coming to rest. Calculate the retarding force

Answer 1

Answer:

Mass of bullet m=10gm=0.01kg

lnitial velocity of bullet u=500 m/s

distance travelled in the wood

s=5cm=0.05 m

v^2 =u^2 +2as,

0=(500)^2-2 a*0.05

a=250000/0.1 a=25*10^5 m/s^2

Retarding force

F=ma=0.01*25*10^5

Retarding force F =25*10^3 N

Work done W=F*s=25*10^3*0.05

Work done W=1250 J

Answer 2

Okay .. first we get the units right .
The mass is given in grams , so we ought to convert it into the SI units
So ,
M=2/1000= 0.002 kg
Initial velocity of bullet u = 400 m/s
Final velocity of the bullet v= 0 m/s (since it is coming to rest )
Displacement of the bullet s = 5 cm or 0.05 m
Applying third equation of motion
v = u + at
V^2-u^2= 2as , we get

0^2 - 400^2= 2a*0.05
0-1600= 0.10a
-1600/0.10=a
-16000 m/s^2 = a
Now,
F=ma from Newton’s Second law of Motion
F= 0.002kg * -16000m/s^2
F=-32N
Negative sign in force implies that the force is acting opposite to the initial direction of the motion of the bullet .
Hope this helps !