A 3.21-kg projectile is fired with an initial speed of 119 m/s at an angle of 32

Question

A 3.21-kg projectile is fired with an initial speed of 119 m/s at an angle of 32 degree with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.94 kg and 2.27 kg. At 3.81 s after the explosion, the 2.27-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 0.94-kg fragment immediately after the explosion. (b) Find the distance between the point of firing and the point at which the 0.94-kg fragment strikes the ground. km (c) Determine the energy released in the explosion. kJ

Explanation / Answer

time takne t reach maximumheight = t = (vo*sintheta)/g = (119*sin32)/9.8 = 6.43 s

maximum height H = vo^2*(sin32)^2 / 2g = (119^2*(sin32)^2)/(2*9.8) = 203 m

horizantal distance travelled before explosion = x = vo*cos32*t = 119*cos32*6.43 = 649 m

inital speed at the maximum height before explosion = u1x = u2x = vo*cos32 = 119*cos32 = 101 m/s

after explosion

m1 = 0.94 kg    m2 = 2.27 kg

t2 = 3.81 s

for v2 along vertical

y = -H

t = 3.81

ay = -9.8 m/s^2

y = v2y*t+0.5*ay*t^2

-203 = v2*3.81 - 0.5*9.8*3.81^2

v2y = -34.6 m/s

v2 = -34.6 j

from momentum conservation

Pf = Pi

m1*v1 + m2*v2y = (m1+m2)*u1x

0.94*v1 - (2.27*34.6)j = 3.21*119*cos32 i

v1 = 344.6i + 83.56 j m/s

++++++++++++

(b)

distance travelled after explosion = x1 = v1x*sqrt*(2H/g)

x1 = 344.6*sqrt(2*203/9.8) = 2218.01 m

total distance from the point of firing = x + x1 = 649+2218.01= 2.867 km

(c)

Ki = 0.5*M*vo^2 = 0.5*3.21*119^2 = 22728.4 J

Kf = 0.5*m1*v1^2 + 0.5*M2*V2^2

kF = (0.5*0.94*(344.6^2+83.56^2)) + (0.5*2.27*34.6^2)

kF = 60452.6 j

energy released = Kf - Ki = 37724.2 J

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