On piercing a distance of 15cm in a wooden block a bullet losses of its initial velocity before coming to rest the further distance travelled by it is
Ans:5cm
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Answered by
6
In first case
Initial velocity becomes Zero
thus
v^2 - u^2 = 2aS
v^2 = 2 a 15
v^2 = 30 a
now for the second case
u = v
thus
and v= 0
thus
0 - v^2 = 2 a x
x = 5 Cm
thus if will further move to 5 Cm
Initial velocity becomes Zero
thus
v^2 - u^2 = 2aS
v^2 = 2 a 15
v^2 = 30 a
now for the second case
u = v
thus
and v= 0
thus
0 - v^2 = 2 a x
x = 5 Cm
thus if will further move to 5 Cm
Answered by
0
Answer:
5cm
Explanation:
from work energy theorem
FS=∆K.E
in S in loss in K.E
15/S=1/2m(v/2)^2-1/2mv^2/0-1/2m(v/2)^
15/S=3v^2/4/v^2/4 =3
S=5cm
I hope this correct answer
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