How much energy is released during the explosion?
A projectile of mass 19.6 kg is fired at an angle of 61.0 degrees above the horizontal and with a speed of 79.0m/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?
Take the free fall acceleration to be g= 9.80m/s^2 .
which i found to be 810m
but i am having trouble with part B
B.How much energy is released during the explosion?
Take the free fall acceleration to be g= 9.80m/s^2.
I know its KE2-KE1 but i havent got the right answer yet
For part B, you have to find the kinetic energy of the system right before and right after the explosion. Since the height of the system right before and after the explosion is the same, there’s no change in potential energy. So it’s all kinetic. Furthermore, the vertical component of velocity is zero at this point right before and after the explosion. So you really only have to contend with the kinetic energy of the system from the horizontal component of velocity right before and after the explosion.
So let’s start. The kinetic energy of the system right before the explosion is this:
KE_pre=0.5*m*[V*cos(theta)]^2
where m=mass of the projectile, V=initial projectile velocity, and theta=angle.
Right after the explosion, you have two pieces, each with 0.5m mass. One piece has zero velocity in either horizontal or vertical direction. The second piece has 2x the horizontal velocity as before and no vertical velocity. You already know this from conserving momentum of the system since you got the first part right.
KE_post_1=0
KE_post_2=0.5*(0.5*m)*[2*V*cos...
KE_post=KE_post_1+KE_post_2
KE_post=m*[V*cos(theta)]^2
The difference is the minimum amount of energy released from the explosion.
E_explosion=KE_post-KE_pre
E_explosion=0.5*m*[V*cos(theta…
Solve for it and you find E_explosion=14375J
That’s the minimum amount of energy that the explosion has to release, just to send the two halves into the paths prescribed by the problem. It may in fact have released more than that. However, anything more than that would have to be in the form of heat, radiation (light), or pressure wave (sound). It can’t be kinetic energy in the projectile halves, otherwise the projectile halves will not follow the paths described by the problem. The problem in part B really should have been written as "what is the minimum amount of energy that must be released during explosion?" in order to give a unique answer. As it is, the answer is anything more than 14375J with the condition that amounts greater than that are released as heat, light, or pressure wave.
February 22nd, 2010 at 10:53 am
For part B, you have to find the kinetic energy of the system right before and right after the explosion. Since the height of the system right before and after the explosion is the same, there’s no change in potential energy. So it’s all kinetic. Furthermore, the vertical component of velocity is zero at this point right before and after the explosion. So you really only have to contend with the kinetic energy of the system from the horizontal component of velocity right before and after the explosion.
So let’s start. The kinetic energy of the system right before the explosion is this:
KE_pre=0.5*m*[V*cos(theta)]^2
where m=mass of the projectile, V=initial projectile velocity, and theta=angle.
Right after the explosion, you have two pieces, each with 0.5m mass. One piece has zero velocity in either horizontal or vertical direction. The second piece has 2x the horizontal velocity as before and no vertical velocity. You already know this from conserving momentum of the system since you got the first part right.
KE_post_1=0
KE_post_2=0.5*(0.5*m)*[2*V*cos...
KE_post=KE_post_1+KE_post_2
KE_post=m*[V*cos(theta)]^2
The difference is the minimum amount of energy released from the explosion.
E_explosion=KE_post-KE_pre
E_explosion=0.5*m*[V*cos(theta…
Solve for it and you find E_explosion=14375J
That’s the minimum amount of energy that the explosion has to release, just to send the two halves into the paths prescribed by the problem. It may in fact have released more than that. However, anything more than that would have to be in the form of heat, radiation (light), or pressure wave (sound). It can’t be kinetic energy in the projectile halves, otherwise the projectile halves will not follow the paths described by the problem. The problem in part B really should have been written as "what is the minimum amount of energy that must be released during explosion?" in order to give a unique answer. As it is, the answer is anything more than 14375J with the condition that amounts greater than that are released as heat, light, or pressure wave.
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