Physics, 24.06.2020 02:01 soccerplayer17

An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 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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Answer from: starreee

K = 3.9 kJ

Explanation:

The kinetic energy ( $K_{T}$ ) added is given by the difference between the final kinetic energy and the initial kinetic energy:

$K_{T} = K_{f} - K_{i}$

The initial kinetic energy is:

$K_{i} = \frac{1}{2}m_{1}v_{1}^{2}$

Where m₁ is the mass of the object before the explosion and v₁ is its velocity

$K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J$

Now, the final kinetic energy is:

$K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2}$

Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces

Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:

$K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0$ (1)

By conservation of momentum we have:

$p_{i} = p_{f}$

$m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3}$

$m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0$

$v_{2} = \frac{5}{4}v_{1}$ (2)

By entering (2) into (1) we have:

$K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J$

Hence, the kinetic energy added is:

$K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J$

Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.

I hope it helps you!

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

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