A 13.0-kg projectile is fired at an angle of 60.0 ° above the horizontal with a
ID: 1438918 • Letter: A
Question
A 13.0-kg projectile is fired at an angle of 60.0° above the horizontal with a speed of 81.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level?km
(b) How much energy is released during the explosion?
J A 13.0-kg projectile is fired at an angle of 60.0° above the horizontal with a speed of 81.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level?
km
(b) How much energy is released during the explosion?
J (a) How far from the point of firing does the other fragment strike if the terrain is level?
km
(b) How much energy is released during the explosion?
J
Explanation / Answer
A. Let + y be upward and +x be horizontal and to the right. Let the two fragments be A and B,
each with mass m. For the projectile before the explosion and the fragments after the explosion.
ax = 0, and ay = -9.81 m/sec^2
Vy^2 = V0y^2 +2a(y - y0)
Given Vy = 0
y - y0 = h = -V0y^2/2*ay = -(81*sin 60 deg)^2/(2*-9.81) = 250.80 m
Just before the explosion the projectile is moving to the right with horizontal velocity
Vx = V0x = 81*cos 60 deg = 40.5 m/sec
After the explosion Vax = 0 since fragment A falls vertically. Conservation of momentum applied to the explosion gives
ma*Vax = mb*Vbx
2*m*40.5 = m*Vbx
Vbx = 81 m/sec
Fragment B has zero initial vertical velocity so
y - y0 = V0y + 0.5*ay*t^2
t = sqrt(-2h/ay) = sqrt(-2*250.80/(-9.81)) = 7.15 sec.
During this time the fragment travels horizontally a distance
d1 = 81*7.15 = 579.15 m
It also took the projectile 7.15 s to travel from launch to maximum height and during this time it travels a horizontal distance of
d2 = 81*cos 60*7.15 = 289.575 m
The second fragment lands 289.575 m + 579.15 m = 868.725 m = 0.868 km from the firing point.
B.
Kf = 0.5*m*v1^2 = 0.5*13*40.5^2 = 10661.625 J
Ki = 0.5*m*u^2 = 0.5*6.5*81^2 = 21323.25 J
dKE = 21323.25 - 10661.625 = 10661.625 J
The energy released in explosion = 10661.625 J
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