Problem (20 points): (Based on 8.108) A 20.0-kg projectile is fired from the ori

Question

Problem (20 points): (Based on 8.108) A 20.0-kg projectile is fired from the origin with an initial velocity Vector v= 40.0 m/s x cap + 69.3m/sy cap. At the highest point of its trajectory, the projectile explodes into two fragments of mass 12.0 kg and 8.00kg. During the brief explosion, the 8.00 kg fragment experiences an impulse Jon8 = 5549 kg m/s x cap + 320.kg m/s y cap (and the 12.0 kg fragment experiences an equal but opposite impulse). At what x-coordinate does each fragment land if the terrain is level? Please use subscripts A, B, C, and D as follows: A=at launch, B=just before exploding, C=just after exploding, D=at landing. A. IDENTIFY Steps A.1 (2 pts) Known and Unknown quantities: Known: Unknown:

Explanation / Answer

at heighest point on horizontal component of velocity will remain and vertical component will get zero.

so in vertical :

0 ^2 - 69.3^2 = 2 x -9.81 x h

h = 244.78 m

so v = 40 i m/s

impulse = change in momentum

on 8 kg :

554 i + 320 j = 8 ( v - 40i)

v = 109.25 i + 40 j m/s

in vertical it have to travel 244.78 m downwrads ,

h = ut + at^2 /2

-244.78 = 40t - 9.81t^2 /2

on solving

t = 12.24 sec

d = ux * t = 109.25 x 12.24 = 1337.22 m

on 12 kg :

-554 i - 320j = 12 (v - 40 i)

v = -6.17 i - 26.67 j m/s

in vertical it have to travel 244.78 m downwrads ,

h = ut + at^2 /2

-244.78 = -26.67t - 9.81t^2 /2

on solving

t = 4.85 sec

d = ux * t = - 6.17x 4.85 = - 29.92 m

for initial position x = 0 then position of explosion ,

time taken to reach max, height

v - u = at

0 - 69.3 = -9.81t

t = 7.06 sec

x = ux*t = 40 x 7.06 =282.57 m

so 8kg : x   =282.57 + 1337.22 =1619.79 m

for 12 kg : x = 282.57 - 29.92 = 252.65 m

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