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A 20.0kg- projectile is fired at an angle of 60.0 above the horizontal with a speed of 80.0 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.How far from the point of firing does the other fragment strike if the terrain is level?
How much energy is released during the explosion?

2007-10-29 04:49:27 · 3 answers · asked by Natiphy2007 1 in Science & Mathematics Physics

3 answers

As the 20.0kg- projectile is fired at an angle of 60.0 above the horizontal with a speed of 80.0 m/s,the horizontal component of its velocity=80 cos 60 = 40 m/s

Initial momentum before exploding = 20*40=800 kgm/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.

mass of each equal part = 10 kg

horizontal momentum of the part that drops is zero as it has no horizontal velocity

Let velocity of other part be V

horizontal momentum of the other part=10V

total final momentum of the two part system in horizontal direction =10V

As no external force has acted momentum is conserved

total final momentum = total initial momentum

10V=800

V=80 m/s

Momentum of second part=800 kgm/s

Thus velocity of second part is twice the velocity of original projectile before explosions

Had the projectile not exploded ,its range R! = (u^2)sin 2O /g=80*80*0.867/g =566.204 m

When it explodes it is at highest point , therefore it has covered half range

When it explodes, it has covered 288.1 m

HAD IT NOT EXPLODED its velocity at highest point would have been 40 m/s and it would have covered 288.1m more, BUT AFTER EXPLOSION ITS VELOCITY AT HIGHEST POINT DOUBLES, hence it will cover double distance that is it will now cover 566.2 m more

Total distance from point of firing=283.10+566.2= 849.3 m
__________________________________________________
KE before exploding=(1/2)MV^2=(1/2)20*40*40=16000 J

KE after exploding=(1/2)mv^2=(1/2)10*80*80=32000 J

Kinetic energy gained=32000 -16000=16000 J

The energy released during the explosion is 16000 J

2007-10-29 05:32:37 · answer #1 · answered by ukmudgal 6 · 8 0

Nowhere near enough information.

2007-10-29 04:54:44 · answer #2 · answered by gebobs 6 · 0 8

I thought it was great!

2016-03-27 16:02:40 · answer #3 · answered by Flake Finder 1 · 0 0

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