A bullet fired into a large block of wood of uniform resistance, hits the block with a velocity of 48 m/s.It penetrates a distance of 60 cm and the velocity is then reduced to 24m/s. It then penetrates a further distance of (in cm)?
Asked by aruni_maiyer
| 11th Jul, 2010,
09:27: PM
Expert Answer:
Let the initial velocity be u and constant retardation be a.
Then
v2 = u2 - 2aS
(24)2 = (48)2 - 2a(60)
or, 120a = 2304 - 576 = 1728
or, a = 14.4 m/s2
Let the bullet covers a total distance of x .
Then using the same acceleration we get:
02 = u2 - 2ax.
0 = 2304 - (2x14.4)(x)
x = 2304/(2x14.4) = 80cm.
So, the bullet covers a total a 80 cms into the target or the bullet goes furthur by 20 cms.
Hope this helps,
Team Topper
Then
v2 = u2 - 2aS
v2 = u2 - 2aS
(24)2 = (48)2 - 2a(60)
or, 120a = 2304 - 576 = 1728
or, a = 14.4 m/s2
Let the bullet covers a total distance of x .
Then using the same acceleration we get:
02 = u2 - 2ax.
0 = 2304 - (2x14.4)(x)
x = 2304/(2x14.4) = 80cm.
or, 120a = 2304 - 576 = 1728
or, a = 14.4 m/s2
Let the bullet covers a total distance of x .
Then using the same acceleration we get:
02 = u2 - 2ax.
0 = 2304 - (2x14.4)(x)
x = 2304/(2x14.4) = 80cm.
So, the bullet covers a total a 80 cms into the target or the bullet goes furthur by 20 cms.
Answered by
| 12th Jul, 2010,
04:11: PM
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