A 0.13-kg firecracker is hanging by a light string from a tree limb. The fuse is
ID: 2302402 • Letter: A
Question
A 0.13-kg firecracker is hanging by a light string from a tree limb. The fuse is lit and the firecracker explodes into three pieces: a small piece (0.01 kg), a medium-sized piece (0.05 kg), and a large piece (0.07 kg). Assume the firecracker is at the origin of a coordinate system such that the +z axis points straight up, the +x axis points due east, and the +y axis points due north. The fragments fly off according to the following momentum vectors:
After the explosion, the component of speed of the smallest piece in both the x and y directions is 5 m/s.
(vSx = vSy = 5 m/s.)
The speed of the largest fragment is 1 m/s in the z direction
(vLz = 1 m/s).
Determine the velocity vector of each fragment immediately after the explosion. (Indicate the direction with the sign of your answer.)
Largest fragment: pL = pLx? ? pLz? Medium fragment: pM = pMy? + pMz? Smallest fragment: pS = ?pSx? ? pSy? m/s) m/s) X m/s) m/s 1.4 m/s 2 m/s 2Explanation / Answer
after explosion
Vsx = 5.....Vsy = 5 m/s...........Vsz =
Vmx = .....Vmy = ..........vmz =
VLx = 0....VLy = 0............Vlz = 1
initial momentumPi = 0
final momentum = Pl +pm +Ps
Pf = Plx X - Plz Z + pmy Y + pmz Z - psx X - psy Y
Pf = (plx - psx) X + (pmy - psy) Y + (pmz - plz ) Z
Pf = (ml*Vlx - ms*vsx) X + (mm*vmy - ms*vsy) Y + (mm*vmz - ml*vlz) Z
Pf = ((0.07*Vlx) - (0.01*5))X + ((0.05*vmy) -(0.01*5)) Y + ((0.05*vmz)-(0.07*1)) Z
Pf = (0.07*vlx -0.05) X + (0.05vmy - 0.05) Y + (0.05vmz - 0.07) Z
pf = Pi
(0.07*vlx -0.05) X + (0.05vmy - 0.05) X + (0.05vmz - 0.07) Z = 0 X + 0 Y + 0Z
equating X
(0.07*vlx -0.05) = 0
vlx = 5/7 = 1.4 m/s
equating Y
(0.05vmy - 0.05) = 0
equating Z
vmy = 1
(0.05vmz - 0.07) = 0
vmz = 0.07/.05 = 7/5 = 1.4 m/s
Vs = 5 X + 5 Y
Vm = 1 Y + 1.4 Z
Vl = 1.4 X + 1 Z
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