Question

b. A bullet of mass 0.4 kg is shot from a rifle with a speed 100 m/s. If the bullet penetrates through a 5 cm wall with a velocity of 20 m/s, what is the force exerted by the wall on the bullet?

Answer 1

U=20m/s

V=0

S=5cm=0.05m

By appliying the formula:V^2-U^2=2×a×s

-400=2×a×(1\20)

a=-4000m/s

m=0.4kg

F=ma

F=0.4×4000

F=1600N

Answer 2

Hey dear,

◆ Answer-

F = -3840 N
.

◆ Explaination-

# Given-

m = 0.4 kg

u = 100 m/s

v = 20 m/s

s = 5 cm = 0.05 m
.

# Solution-

Using Newton's third kinematic eqn -

v^2 = u^2 + 2as

a = (v^2-u^2) / 2s

a = (20^2-100^2) / (2×0.05)

a = (400-10000) / (0.1)

a = -9600 N/m

Force exerted by the wall on bullet is

F = ma

F = -0.4×9600

F = -3840 N

Force exerted by the wall on bullet is 3840 N in opposite direction.

Hope this helps...