Can anyone help me with this physics problem? It seems very simple but I can't get it.
An object, with mass 36 kg and speed 14 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?
An object, with mass 36 kg and speed 14 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?
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In the frame of the centre of mass the light piece has a velocity of 14 m/s in one direction.
And the heavy piece has a velocity of 7 m/s in the other direction.
The heavy piece weighs 24 kg
the light piece weighs 12 kg
So the energy given by the explosion is 1/2 * 24 * 7 ^2 +1/2 * 12 * 14^2
= 1764 joules.
this must also be true from the position of any other observer.
But to demonstrate.
The original kinetic energy of the system measured from the observer was
1/2 m v^2 = 1/2 * 36 * 14 ^2
The final kinetic energy is 1/2 * 24 * 21^2
hence the gain is = 1/2 *(24 * 21^2 - 36*14^2 )
= 1764 J
So no matter what the inertial observer the change in energy must always be in agreement.
and the final
And the heavy piece has a velocity of 7 m/s in the other direction.
The heavy piece weighs 24 kg
the light piece weighs 12 kg
So the energy given by the explosion is 1/2 * 24 * 7 ^2 +1/2 * 12 * 14^2
= 1764 joules.
this must also be true from the position of any other observer.
But to demonstrate.
The original kinetic energy of the system measured from the observer was
1/2 m v^2 = 1/2 * 36 * 14 ^2
The final kinetic energy is 1/2 * 24 * 21^2
hence the gain is = 1/2 *(24 * 21^2 - 36*14^2 )
= 1764 J
So no matter what the inertial observer the change in energy must always be in agreement.
and the final
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first, we calculate the speed of the second piece using conservation of momentum:
36 (14) = 12 (0) + 24 (v) => v = 21 m/s
we now calculate the KE added by subtracting the KE of the object before the explosion from the KE after the explosion:
(24*(21)^2) /2 - (36*(14)^2) /2 = 1764 J
36 (14) = 12 (0) + 24 (v) => v = 21 m/s
we now calculate the KE added by subtracting the KE of the object before the explosion from the KE after the explosion:
(24*(21)^2) /2 - (36*(14)^2) /2 = 1764 J
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K E added = kinetic energy lost by the body = 1/2 m v^2 = 1/2 x 36 x 14 x14 = 3528 J.