For a ballistics study, a 2.2 gm bullet is fired into soft wood. Thebullet strikes the wood surface with a velocity of 370 m/s andpenetrates 0.17 m. Find (a) the constant retarding force, N, (b)the deceleration, m/s^2, (c) the time required to stop the bullet.
m=2.2 gm=2.2×10−3 kgv=370 m/sd=0.17 mm = 2.2\textsf{ gm}= 2.2×10^{-3}\textsf{ kg} \\ v = 370\textsf{ m/s}\\ d = 0.17\textsf{ m}\\m=2.2 gm=2.2×10−3 kgv=370 m/sd=0.17 m
(a)
Fd=EkFd=1/2mv2Fd = E_k\\ F d= 1/2mv²Fd=EkFd=1/2mv2
F=mv22d=2.2×10−3×3702×0.17F = \dfrac{mv²}{2d} = \dfrac{2.2×10^{-3}× 370}{2×0.17}F=2dmv2=2×0.172.2×10−3×370
=2.39 N= 2.39\textsf{ N}=2.39 N
(b)
F=ma2.39=2.2×10−3×aa=1086.5 ms−2F = ma\\ 2.39 = 2.2×10^{-3}× a\\ a = 1086.5\textsf{ ms}^{-2}F=ma2.39=2.2×10−3×aa=1086.5 ms−2
(c)
Ft=mv22.39×t=2.2×10−3×(3702)t=126.02 sFt = mv²\\ 2.39×t = 2.2× 10^{-3}× (370²)\\ t = 126.02\textsf{ s}Ft=mv22.39×t=2.2×10−3×(3702)t=126.02 s
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