bullet of 10 g strikes a sand bag at a speed of 103 m/s and gets
embedded after travelling 5 cm.
Calculate
(I)The resistive force exerted by the sand on the bullet
(ii) The time taken by the bullet to come to rest.
Answers
Mass of bullet should be =10gm=0.01kg
initial velocity (U)of bullet=1000m/s
final velocity (V)=0m/s(because it has stopped after striking sand)
displacement of bullet=5cm=0.05m
a)using 2as=V ²-U²
2(a)(0.05)=0²-1000²
2a(5/100)=-1000000
2a(1/20)=-1000000
a(1/10)=-1000000
a=-10000000=-10power7
acceleration of bullet is -10power7or -10000000
by using F=ma
F=0.01(10000000)
F=1/100(10000000)
F=-100000N
there fore resistive force of bullet =-100000N
b)by using a=v-u/t
-10000000=0-1000/t
-10000000=-1000/t
-10000=t
or
t=-10⁴seconds
hope it helps you
Answer:
(a)Resistive force exerted by sand on bullet.
Mass of bullet should be =10gm=0.01kg
initial velocity (U)of bullet=1000m/s
final velocity (V)=0m/s(because it has stopped after striking sand)
displacement of bullet=5cm=0.05m
a)using 2as=V ²-U²
2(a)(0.05)=0²-1000²
2a(5/100)=-1000000
2a(1/20)=-1000000
a(1/10)=-1000000
a=-10000000=-10^7
acceleration of bullet is -10^7=-10000000
by using F=ma
F=0.01(10000000)
F=1/100(10000000)
F=-100000N
there fore resistive force of bullet =-100000N
(b)Time taken by bullet to come to rest.
Given mass of bullet m = 10 g = 0.01 kg
Speed v = 1000 m/s
Distance s = 5 cm = 0.05 m
We know that F = ma
and v^2 = u^2 – 2as
or a = v^2 – u^2 / 2s
Now F = m (v^2 – u^2 / 2s)
= 0.01 x (1000^2 – 0 / 2 x 0.05)
Resistive force will be F = 10^5 N
Now time taken to come to rest will be
S = (u + v / 2) t
So t = 2s / u + v
= 2 x 0.05 / 1000
t = 10 ^-4 secs