a bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03s. calculate the distance of penetration of the bullet into the block? can any one help me with this
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Answered by
4
vf = vi + at
0 = 150 m/s + a(0.03 s)
a = -5000 m/s^2
vf^2 = vi^2 + 2ad
0 = (150 m/s)^2 + 2(-5000 m/s^2)(d)
d = 2.25 m
F = ma
F = (0.01 kg)(5000 m/s^2)
F = 50 N
0 = 150 m/s + a(0.03 s)
a = -5000 m/s^2
vf^2 = vi^2 + 2ad
0 = (150 m/s)^2 + 2(-5000 m/s^2)(d)
d = 2.25 m
F = ma
F = (0.01 kg)(5000 m/s^2)
F = 50 N
Answered by
2
v=u+at
0=150+a*0.03
-150/0.03=a
a=-5000m/s²
s=ut+1/2 at²
s=150*0.03-0.5*5000*0.03*0.03
s=4.5-2.25
s=2.25m
The distance penetrated is 2.25m
0=150+a*0.03
-150/0.03=a
a=-5000m/s²
s=ut+1/2 at²
s=150*0.03-0.5*5000*0.03*0.03
s=4.5-2.25
s=2.25m
The distance penetrated is 2.25m
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