Question

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?

Answer 1

Answer:

The motion of the projectile can be split up in two motions, the horizontal and the vertical motion. The vertical velocity v for a time t, acceleration g and initial velocity is given by:

(1) $v_y=v_0sin\theta-gt$

At the highest point of flight the velocity will be zero and equation 1 can be solved for time t:

(2) $t=\frac{v_0sin\theta}{g}$

The velocity of the projectile in the horizontal direction up to the highest point is constant:

(3) $v_x=v_0cos\theta$

At the highest point of flight the projectile explodes. All forces are internal, so momentum must be conserved.

The momentum p before the explosion:

(4) $p=mv_x$

The momentum after the explosion with half the projectile having zero velocity:

(5) $p=\frac{m}{2}v_x'$

The velocity of the rest of the projectile after the explosion can be found from equations 3, 4 and 5:

(6) $v_x'=2v_x=2v_0cos\theta$

The total distance of the projectile fragment can now be calculated:

(7) $x=v_xt+v_x't=3v_0cos(\theta) t=3v_0cos(\theta)\frac{v_0sin\theta}{g}=\frac{3v_0^2}{2g}sin(2\theta)$

$x=847.5m$

The explosion releases kinetic energy. The kinetic energy of the projectile before the explosion:

(8) $E_1=\frac{1}{2}mv_x^2$

The kinetic energy after the explosion:

(9) $E_2=\frac{1}{2}\frac{m}{2}(2v_x)^2=mv_x^2$

The energy released:

(10) $E=E_2-E_1=\frac{1}{2}mv_x^2=\frac{1}{2}m(v_0cos\theta)^2$

$E=16000J$