Question

a bullet of 20g strikes a sand bag at a speed of 200m/s and gets embedded after travelling 2cm then the resistiue
force is​

Answer 1

Answer

(b) Given, u = 200 m/s, v = 0, s = 0.02 m

and m = 20 g = 20 x 10-3 kg

From third equation of motion, v2 = u2 + 2as a

⇒ a =

v

2

−

u

2

2

s

=

(

0

)

2

−

(

200

)

2

2

X

0.02

= -106 m/s2

Negative sign indicates retardation.

.: Resistive force, F = ma = 20 x 10-3 x 106

= 2 x 104 N

Answer 2

$\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\red{W}\green{E} \purple{R}}}★$

$Given : -$

$mass \: of \: the bul \: let (m) = 0.02 kg$

$initial vel \: ocity of \: the bul \: let (u) \\ = 200 m/s$

$final vel \: ocity of \: the bu \: llet (v) \\ = 0 m/s$

$distance tra \: versed by \: the bull \: \\ et in t \: he san \: d before co \: mi \\ ng to rest (S) = 0.03 m$

$From \: the \: equation \: of \: motion,$

$v2–u2=2aS$

$⇒0−2002=2aS$

$⇒a=−32×106 (Here a is \: the \: deceleration)$

$F=m(−a)$

$⇒F=0.02×32×106$

$Note: 1N=105dyne$

$⇒F=13.3×108dyne$