Question

Using second law of motion , derive the relation between force and acceleration . A bullet of 10 g strikes a sand bag at a speed o 10^3m/s and gets embedded after travelling 5 cm . Calculate :

(A) The resistance force exerted by the sand on the bullet.

(B) The time taken by the bullet come to rest

Answer 1

Second law of motion tells that change in momentum per unit time is directly proportional to force
hence
F = Kmv-mu/t
F=Km(v-u)/t
F=Kma
F = ma (k= 1)
Here s= 5/100
u=10^3
v= 0
${v}^{2} = {u}^{2} + 2as$
$0 = {10}^{6} + 2 \times \frac{5}{100} \times a$
${ - 10}^{6} = \frac{1}{10} a \\ a = - {10}^{7}$
m=10/10^3
F= ma
$f = \frac{10}{ {10}^{3} } \times { - 10}^{7} \\ f = - {10}^{5}$
v=u+at
0= 10^3+-10^7t
-10^3=-10^7t
-10^3\-10^7=t
t=10^-4
HOPE IT HELPS AND IF POSSIBLE THEN PLEASE MARK IT AS THE BRAINLIEST ANSWER.

Answer 2

Answer:

Explanation:

Second law of motion tells that change in momentum per unit time is directly proportional to force

hence

F = Kmv-mu/t

F=Km(v-u)/t

F=Kma

F = ma (k= 1)

Here s= 5/100

u=10^3

v= 0

m=10/10^3

F= ma

v=u+at

0= 10^3+-10^7t

-10^3=-10^7t

-10^3\-10^7=t

t=10^-4

HOPE IT HELPS AND IF POSSIBLE THEN PLEASE MARK IT AS THE BRAINLIEST ANSWER.