A 20.0-kg projectile is fired from the origin with an initial velocity v = 40.0m

Question

A 20.0-kg projectile is fired from the origin with an initial velocity v = 40.0m/sx + 69.3m/sy. 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 = 554kgm/sx + 320.kgm/sy (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.

Explanation / Answer

At the launch , the coordinate of the projectile is given as

Xa =0 Ya=0

initial velocity of projectile in X-direction = Vax = 40 m/s

initial velocity of projectile in Y-direction = Vay = 69.3 m/s

At the highest point before exploding

velocity in X-direction = 40 m/s

velocity in Y-direction = 0 m/s

Using kinematics equation along Y-direction ::

Vf2 = Vi2 + 2 a d

0 = 69.3^2 + 2 (-9.8) d

d= 245.03 m

Vf = Vi + at

0 = 69.3 + (-9.8) t

t = 7.07

X = V t = 40 x 7.1 = 284 m

So coordinates of the highest point are X = 284 and Y = 245.03

at the highest point projectile explodes

for 8 kg fragment ::

impulse along X-direction = 554 kgm/s

m (Vf - Vi) = 554

8(Vf - 40) = 554

Vf =109.25 m/s velocity of 8 kg fragment after explosion in X-direction

impulse along Y-direction = 320 kgm/s

m (Vf - Vi) = 320

8(Vf - 0) = 320

Vf =40 m/s velocity of 8 kg fragment after explosion in Y-direction

Using kinematics equation along Y-direction::

d = Vi t + (0.5) a t^2

245.03 = 40 t + 4.9 t^2

t = 4.1 sec

X = v t = (109.25)(4.1 sec)

X = 447.93 m

8 kg fragment lands at X = 447.93 m

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