An object with mass m, initially at rest, explodes into two...

An object with mass $m,$ initially at rest, explodes into two fragments, one with mass $m_{A}$ and the other with mass $m_{B},$ where $m_{A}+m_{B}=m .$ (a) If energy $Q$ is released in the explosion, how much kinetic energy does each fragment have immediately after the collision? (b) What percentage of the total energy released does each fragment get when one fragment has four times the mass of the other?

Key Concepts

Conservation of Momentum

In any explosion, the system is usually isolated with no external forces acting on it. As a result, the total momentum before and after the explosion remains the same. When an object at rest explodes, the momentum of the resulting fragments must sum to zero, dictating a precise relationship between their velocities and masses.

Energy Transformation and Conservation

Explosions release stored internal energy which is converted into the kinetic energy of the fragments. Despite the energy conversion, the overall energy is conserved within the system, provided all forms are accounted for. The partitioning of this released energy into kinetic energy among the fragments follows from conservation laws and the dynamics of the explosion.

Kinetic Energy Partitioning

The distribution of kinetic energy among the fragments depends on their masses as well as their velocities. Due to momentum conservation, lighter fragments typically attain higher speeds, potentially acquiring a larger share of the kinetic energy relative to heavier fragments. This inverse relationship between the fragment mass and its kinetic energy is crucial in understanding energy distribution in collision and explosion scenarios.

Center-of-Mass Frame Analysis

Analyzing the problem in the center-of-mass frame simplifies the calculations, particularly because the overall motion is reduced to relative motions of the individual fragments. In this frame, the symmetry of the situation can be exploited, allowing for a clearer derivation of the relationship between fragment velocities and masses, and provides a straightforward way to determine each fragment's kinetic energy.

Transcript

00:01 There is no external horizontal force on the object so momentum will be conserved so we can write pa plus pb equals to zero so from here we can write p a equals to minus pb so now putting values we get p a va va va m a equals to minus of minus mb vb okay so from here we can write vb equals to m a divided by v a m a divided by m b multiplied by v a okay say this is equation number one now the total energy released from the explosion is the kinetic energy so we can write q equals to total kinetic energy that is kinetic energy of a plus kinetic energy of b okay so now we can write q equals to k a plus 1 by 2 mb vb square now putting the value of vb in this equation we get q equals to k a plus 1 by 2 mb and vb is m a divided by m b multiplied by m b multiplied by v a and whole square so now after solving it we get q equals to and putting value of k a that is half m a v a square half m a v a square plus one by two and solving dates we guess m a va va square m a by mv this this is the relation so from here by taking common we get 1 by 2 m a v a whole sphere and we get 1 plus m a divided by mb okay so we can write this is k so we can write q equals to k a m a m a plus m b divided by mb okay so from here kinetic energy of a comes out to be mb m b divided by m a plus m b multiplied by q so this is the answer for the kinetic energy of particle a okay so now moving forward substituting this equation we get q equals to m b divided by m a plus m b multiplied by q plus kb okay so now from here we can write kb equals to 1 minus m b divided by m a plus m b multiplied by q so from here after solving it kb equals to m a divided by m a plus m b multiplied by q so this is the kinetic and of b in the equation form now for the part b we have some values that is we have m a equals to 4 times of mb so now putting all these equations we get k a equals to m b divided by m a is 4 times of mb so 4 mb plus m b multiplied by q so from here we get q by 5.

04:00 So from here now we can write ka divided by q equals to 20 .0 % this is the answer for this part.

04:13 So similarly we can get kb equals to 100 % minus ka by q.

04:26 That is 80%...

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