A bullet is fired from a rifle emerging from the muzzle at 349 meters per second it strikes a sandbag some distance away having lost 10 percent of its velocity due to air resistance if it penetrates the sand bag to a depth of 12.0 centimetres how long did it take for the bullet to come to rest in the sandbag?
Answers
Answer:
Answer:
The force imparted on the bullet is 0.9 Ns
Explanation:
The force given is
F=600-2\times 10^5tF=600−2×10
5
t
If the force is zero then
0=600-2\times 10^5t0=600−2×10
5
t
\implies 2\times 10^5t=600⟹2×10
5
t=600
\implies t=300\times 10^{-5}⟹t=300×10
−5
\implies t=0.003⟹t=0.003 seconds
The impulse due to force F is given by
\boxed{I=\int Fdt}
I=∫Fdt
Therefore, Impulse imparted to the bullet
=\int_0^{0.003} Fdt=∫
0
0.003
Fdt
=\int_0^{0.003} [600-2\times 10^5t]dt=∫
0
0.003
[600−2×10
5
t]dt
=\int_0^{0.003} 600dt-2\times 10^5\int_0^{0.003}tdt=∫
0
0.003
600dt−2×10
5
∫
0
0.003
tdt
=600\int_0^{0.003}dt-2\times 10^5\int_0^{0.003}tdt=600∫
0
0.003
dt−2×10
5
∫
0
0.003
tdt
=600t\Bigr|_0^{0.003}-2\times 10^5(t^2/2)\Bigr|_0^{0.003}=600t
∣
∣
∣
∣
0
0.003
−2×10
5
(t
2
/2)
∣
∣
∣
∣
0
0.003
=1.8-0.9=1.8−0.9
=0.9=0.9 Ns