An object, with mass 55 kg and speed 19 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?
dang. The smaller piece has v=0. Make the adjustment.
To calculate the kinetic energy added to the system during the explosion, you need to find the change in kinetic energy of the objects involved.
First, let's find the initial kinetic energy of the object before the explosion:
Kinetic energy = (1/2) * mass * (speed^2)
= (1/2) * 55 kg * (19 m/s)^2
= 10247.5 J
Now, let's consider the two pieces after the explosion. Let the mass of the less massive piece be 'm' and the mass of the more massive piece be '2m'.
Since the less massive piece stops relative to the observer, its final velocity is 0 m/s.
The initial kinetic energy of the system is equal to the sum of the kinetic energies of the two pieces:
Initial kinetic energy = (1/2) * m * (speed^2) + (1/2) * (2m) * (speed^2)
= (1/2) * m * (19 m/s)^2 + (1/2) * (2m) * (19 m/s)^2
= 361 m^2 J
Since the less massive piece stops completely, all of the initial kinetic energy is transferred to the more massive piece. Therefore, the change in kinetic energy is:
Change in kinetic energy = Initial kinetic energy - Final kinetic energy
= 10247.5 J - 0 J
= 10247.5 J
Therefore, the kinetic energy added to the system during the explosion, as measured in the observer's frame of reference, is 10247.5 J.
find v of the greater piece by conserving momentum:
55*19 = 55/3 * -19 + 110/3 v
actually, because each term has a factor of 55, we can use
19 = -19/3 + 2v/3
Now, having all the speeds, figure the new KE = 1/2 mv^2 values