A bullet moving with a speed of 150 m s⁻¹ strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Answers
Answered by
61
Hii dear
# Given-
u=150m/s
v=0m/s
s=3.5cm=3.5×10^-2m
# Solution-
By 3rd kinematic eqn,
v^2 = u^2 + 2as
0^2 = 150^2 + 2×3.5×10^-2×a
a = -3.2×10^5m/s^2
By 1st kinematic eqn,
v = u + at
t = (v-u)/a
t = (0-150)/(-3.2×10^5)
t = 4.69×10^-4 s
Hence, bullet will have retardation of -3.2×10^5 m/s^2 and will take 4.7×10^-4 s to penetrate.
Hope this helps you...
# Given-
u=150m/s
v=0m/s
s=3.5cm=3.5×10^-2m
# Solution-
By 3rd kinematic eqn,
v^2 = u^2 + 2as
0^2 = 150^2 + 2×3.5×10^-2×a
a = -3.2×10^5m/s^2
By 1st kinematic eqn,
v = u + at
t = (v-u)/a
t = (0-150)/(-3.2×10^5)
t = 4.69×10^-4 s
Hence, bullet will have retardation of -3.2×10^5 m/s^2 and will take 4.7×10^-4 s to penetrate.
Hope this helps you...
Answered by
30
Given :
U=150 m/s
S=3.5cm=0.035m
V=0 m/s
From third equation of motion:
v²-u²=2as
a=v²-u²/2s
=o -150x-150/2x0.035
=-3.214 x 10⁵ m/s2
= - 3.214 x 10⁵ m/s2
T=v-u/a= 0- 150 / -3.214 x 10⁵
=-150/ 3.214 x 10⁵
=4.67 x 10⁻⁴ sec
∴ The magnitude of its retardation is 3.214 x 10⁵ m/s2 and time taken for it to stop after striking the tree is 4.67 x 10⁻⁴ sec
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