A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the magnitude of the force exerted by the wooden block on the bullet.
Answers
Answer:Given, Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg
Initial velocity of bullet, u = 150 m/s
Since bullet comes to rest, thus final velocity, v =0
Time, t = 0.03 s
Distance of penetration, i.e. Distance, covered (s)=?
Magnitude of force exerted by wooden block =?
We know that,
v = u + a t
⇒
0
=
150
m
s
−
1
+
a
×
0.03
s
⇒
−
150
m
/
s
=
a
×
0.03
s
⇒
a
=
−
150
m
/
s
0.03
- s
=
−
5000
m
s
−
2We know that,
s
=
u
t
+
1
2
a
t
2
⇒
s
=
150
m
/
s
×
0.03
s
+
1
2
(
−
5000
m
s
−
2
)
×
(
0.03
s
)
2
⇒
s
=
4.5
m
−
2500
m
s
−
2
×
0.0009
s
2
⇒
s
=
4.5
m
−
2.25
m
⇒
s
=
2.25
m
Magnitude of force exerted by wooden block
We know that, Force = mass x acceleration
Or,
F
=
0.01
k
g
×
–
5000
m
s
−
2
=
−
50
N
Therefore, Penetration of bullet in wooden block = 2.25 m
Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.
Answer:
50 N
Explaination
= given,mass=10g=0.010kg
v=150m/s
u=0m/s
t=0.03s
F=m×a
=m×(v-u)/t
=0.010kg×(150m/s-0m/s)/0.03s
=1kg×150m/s. /3s
= 50kgm/S2
= 50 Newton