A 0.13-kg firecracker is hanging by a light string from a tree limb. The fuse is

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 (v_? = 1 m/s). Determine the velocity vector of each fragment immediately after the explosion. (Indicate the direction with the sign of your answer.)

Explanation / 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 i - Plzk + pmy j + pmz k + psxi - psy j

Pf = plx + psx) i + (pmy - psy) j + (pmz - plz ) k

Pf = (ml*Vlx + ms*vsx) i + (mm*vmy - ms*vsy)j + (mm*vmz - ml*vlz)k

Pf = ((0.07*0) + (0.01*5))i + ((0.05*vmyj) -(0.01*5))j + ((0.05*vmz)-(0.07*1))k

Pf = 0.05 i + (0.05vmy - 0.05) j + (0.05vmz - 0.07)k

pf = Pi

0.05 i + (0.05vmy - 0.05) j + (0.05vmz - 0.07)k = 0

(0.05vmy - 0.05) = 0

vmy = 1

(0.05vmz - 0.07) = 0

vmz = 0.07/.05 = 7/5 = 1.4 m/s

Vs = 5 i + 5 j

Vm = 1 j + 1.4 k

Vl = 0 i + 1 k

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