Question

a bullet of mass 10g travelling horizontally with a velocity 150ms to the power-1 strikes a stationary wooden blog and comes to rest in 0.03s. calculate the distance of penetration of bullet. also calculate magnitude of force

Answer 1

Explanation:

M=0.01Kg

u=150m/s

v=0

t=0.03s

since it is decelerating

a=v-u/t

a=0-150/0.03

a=(-150)/0.03=(-5000)m/s²

so the bullet is coming to rest it will apply a

negative force to the plank

and the plank will apply a positive resistance

to the bullet by newtons third law of motion

-F(A,B)=+F(B,A)

where A is the bullet abd B is the plank

So

F=ma=5000×10/1000=50N resistance force by

the plank

distance travelled =

v²=u²+2as

0²=150²-2×5000×s

2×5000×s=150×150

S=2.25m