Question

6. A 0.09 kg arrow hits a target at 22 m/s and penetrates 4 cm before stopping.

Answer 1

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