A projectile with a mass of 15 kg is fired from point O, with a velocity u 300 m

Question

A projectile with a mass of 15 kg is fired from point O, with a velocity u 300 m/s in the vertical x- plane at the inclination shown. When it reaches the top of its trajectory at P, it explodes into three tragments A, B, and C. Immediately after the explosion, fragment A is observed to rise vertically a distance of 450 m above P, and fragment 8 is seen to have a horizontal velocity va and eventually lands at point Q. When recovered, the masses of the fragments A, B, and C are found to be 3, 7, and 5 kg, respectively. Calculate the velocity which fragment C has immediately after the explosion work (10 points) and vector solution (5 points)- 15 points vc 4000 m-

Explanation / Answer

form the given data

mass of projectile. M = 15 kg

u = 300 m/s

theta = arctan(4/3)

sin(theta) = 4/5

cos(theta) = 3/5

when it reaches the top

befire expolosion, its speed along x axis is ucos(theta) = 300*3/5 = 180 m/s

after explosion

A rises vertically upwards, hmax = 450 m

lets its speed be va

then 2*g*hmax = va^2

va = sqrt(2g*hmax) = 93.96275858 m/s

B moves horizontally with initial speed along x axis ux, along y axis uy

let velocity of fragment C be Vx i + Vy j + Vz k

then from conservation of momentum

ma =3 kg

mb = 7 kg

mc = 5 kg

Mucos(theta) i = ma*va*k + mb(uxi + uyj) + mc(Vxi + Vyj + Vzk)

also,

ux*t = 4000*cos(45)

ux = uy

also

h = 0.5*g*t^2

now,

2*g*h = (usin(theta))^2

h = 2935.779816 m

hence

t = 24.464831804281 s

hence

ux = uy = 115.61195872400 m/s

hence comparing

15*300*3/5 = 7*115.61195872400 + 5*Vx

0 = 7*115.611958 + 5*Vy

0 = 3* 93.96275858+ 5*Vz

hence

Vx = 378.1432 m/s

Vy = -161.856741 m/s

Vz = -56.37765514 m/s

hence

V = 378.1432i +- 161.856j - 56.377k m/s

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