6. a 0.09 kg arrow hits a target at 22 m/s and penetrates 4 cm before stopping.
a. with what average force did the target exert on the arrow?
b. with what average force did the arrow exert on the target?
c. if the speed was doubled, but the forces remained the same, what’s the new penetration depth?

A ) v = v o - a t
0 = 22 - a · t
a · t = 22
d = v o · t - a t²/2
0.04 = 22 t - 22 t / 2
0.04 = 11 t
t = 0.04 : 11 = 0.003636 s
a = 22 / t
a = 6050 m/s²
F = m · a = 0.09 kg · 6050 m/s²
F ( target→arrow) = - 544.5 N
b ) F ( arrow→target ) = 544.5 N
c )  If the speed was doubled:  v = 44 m/s;
F = a m
a = 6050 m/s²
a · t = 44
t = 6050 : 0.04
t = 0.007272 s
d = 44 t - 44 t/2 = 22 t
d = 22 · 0.007272
d = 0.16 m = 16 cm

Transfer of electrons. hope it !