A bullet of mass 10 g strikes a wooden block with a velocity of 300 m/s after penetrating 20cm into it iys velocity drops to 200 m/s then average resistance offered by the block is
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mass of bullet , m = 10g = 0.01kg
initial velocity, u = 300m/s
final velocity , v = 200m/s
distance travelled , s = 20cm = 0.2m
use formula, v² = u² + 2as
(200)² = (300)² + 2a(0.2)
400 - 900 = 0.4a
-500 = 0.4a
a = -1250 m/s²
resistance offered by the block = - force exerted by bullet [ from Newton's 3rd law ]
= - mass of bullet × retardation
= -0.01 × -1250 N
= 12.50N
hence, resistance offered by the block = 12.5N
initial velocity, u = 300m/s
final velocity , v = 200m/s
distance travelled , s = 20cm = 0.2m
use formula, v² = u² + 2as
(200)² = (300)² + 2a(0.2)
400 - 900 = 0.4a
-500 = 0.4a
a = -1250 m/s²
resistance offered by the block = - force exerted by bullet [ from Newton's 3rd law ]
= - mass of bullet × retardation
= -0.01 × -1250 N
= 12.50N
hence, resistance offered by the block = 12.5N
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Answer:
F=1250N
Explanation:
given in that picture please check it
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