Question

A lead bullet of mass 20 gm travelling with a velocity 350 m/s comes to rest after pentrating 40 cm in a still target .Find resistive force offered by target

Answer 1

hello.... ☺


Mass of the bullet, m
= 20 g
..............................or.......................................
= 0.02 kg

Initial velocity, u
= 350 m/s

Final velocity, v
= 0

Distance ,S
= 40 cm
= 0.4 m

Using v2 = u2 + 2aS

=0 = 3502 + 2(a)(0.4)

= a = -153125 m/s2

The negative sign comes because acceleration opposes the motion.......

thank you ☺