Two fragments of mass $m$ and $2m$ move along mutually perpendicular directions with speeds $2v$ and $v$ respectively.
The energy released during the process is,
Understanding Particle Decay and Energy Release
This problem involves a particle undergoing spontaneous decay into multiple fragments. We need to determine the total kinetic energy released during this process. The key physics principles involved are the conservation of linear momentum and the concept of energy release, which relates to the change in kinetic energy.
Analyzing the Initial State
Before the decay, the particle has a total mass of $6m$ and is at rest. This means:
- Initial Mass: $M = 6m$
- Initial Velocity: $V_i = 0$
- Initial Momentum: $P_i = M \times V_i = 6m \times 0 = 0$
- Initial Kinetic Energy: $KE_i = \frac{1}{2} M V_i^2 = \frac{1}{2} (6m) (0)^2 = 0$
Since the particle starts at rest, its initial momentum and kinetic energy are both zero.
Conservation of Momentum During Decay
The particle breaks into three fragments. According to the conservation of linear momentum, the total momentum of the system must remain constant because there are no external forces acting on the particle as it breaks apart on its own.
Let the three fragments have masses $m_1$, $m_2$, $m_3$ and velocities $\vec{v}_1$, $\vec{v}_2$, $\vec{v}_3$ respectively. The total final momentum $P_f$ is the vector sum of the momenta of the fragments:
$ P_f = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3 $Since momentum must be conserved ($P_i = P_f$):
$ 0 = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3 $Determining the Properties of the Fragments
We are given the details for the first two fragments:
- Fragment 1: Mass $m_1 = m$, Speed $|\vec{v}_1| = 2v$. Let's assume it moves along the x-axis, so $\vec{v}_1 = 2v \hat{i}$.
- Fragment 2: Mass $m_2 = 2m$, Speed $|\vec{v}_2| = v$. It moves along a direction perpendicular to the first fragment. Let's assume it moves along the y-axis, so $\vec{v}_2 = v \hat{j}$.
The mass of the third fragment ($m_3$) can be found by subtracting the masses of the first two fragments from the total initial mass:
$ m_3 = (\text{Initial Mass}) - m_1 - m_2 $ $ m_3 = 6m - m - 2m = 3m $Calculating the Velocity of the Third Fragment
Using the conservation of momentum equation:
$ 0 = m(2v \hat{i}) + 2m(v \hat{j}) + 3m\vec{v}_3 $Rearranging to solve for the momentum of the third fragment ($3m\vec{v}_3$):
$ 3m\vec{v}_3 = -2mv \hat{i} - 2mv \hat{j} $Now, we can find the velocity vector $\vec{v}_3$ by dividing by the mass $m_3 = 3m$:
$ \vec{v}_3 = \frac{-2mv \hat{i} - 2mv \hat{j}}{3m} $ $ \vec{v}_3 = -\frac{2}{3}v \hat{i} - \frac{2}{3}v \hat{j} $This is the velocity vector of the third fragment. To calculate its kinetic energy, we need its speed (the magnitude of $\vec{v}_3$).
$ |\vec{v}_3| = \sqrt{\left(-\frac{2}{3}v\right)^2 + \left(-\frac{2}{3}v\right)^2} $ $ |\vec{v}_3| = \sqrt{\frac{4}{9}v^2 + \frac{4}{9}v^2} = \sqrt{\frac{8}{9}v^2} = \frac{2\sqrt{2}}{3}v $Calculating the Final Kinetic Energy
The total kinetic energy after the decay ($KE_f$) is the sum of the kinetic energies of the three fragments.
Kinetic Energy of Fragment 1 ($KE_1$):
$ KE_1 = \frac{1}{2} m_1 |\vec{v}_1|^2 = \frac{1}{2} (m) (2v)^2 = \frac{1}{2} m (4v^2) = 2mv^2 $Kinetic Energy of Fragment 2 ($KE_2$):
$ KE_2 = \frac{1}{2} m_2 |\vec{v}_2|^2 = \frac{1}{2} (2m) (v)^2 = mv^2 $Kinetic Energy of Fragment 3 ($KE_3$):
$ KE_3 = \frac{1}{2} m_3 |\vec{v}_3|^2 = \frac{1}{2} (3m) \left(\frac{2\sqrt{2}}{3}v\right)^2 $ $ KE_3 = \frac{1}{2} (3m) \left(\frac{8}{9}v^2\right) = \frac{1}{2} m \left(\frac{24}{9}v^2\right) = \frac{1}{2} m \left(\frac{8}{3}v^2\right) = \frac{4}{3}mv^2 $Total Final Kinetic Energy ($KE_f$):
$ KE_f = KE_1 + KE_2 + KE_3 $ $ KE_f = 2mv^2 + mv^2 + \frac{4}{3}mv^2 $To add these terms, find a common denominator:
$ KE_f = \frac{6}{3}mv^2 + \frac{3}{3}mv^2 + \frac{4}{3}mv^2 $ $ KE_f = \left(\frac{6 + 3 + 4}{3}\right)mv^2 = \frac{13}{3}mv^2 $Determining the Energy Released
The energy released during the process is the difference between the total final kinetic energy and the total initial kinetic energy.
$ \text{Energy Released} = KE_f - KE_i $ $ \text{Energy Released} = \frac{13}{3}mv^2 - 0 $ $ \text{Energy Released} = \frac{13}{3}mv^2 $This calculated value represents the energy converted from the internal potential energy of the particle into the kinetic energy of the fragments during the decay.
Important Questions from Conservation of Linear Momentum
- Body A of mass $m$ moving with speed $u$ collides with another body B of mass $3m$, at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is:
What are the dimensions of angular momentum?
The propulsion of a rocket is based on which of Newton's laws of motion?
For a system of interacting particles, which condition is fundamental for the conservation of its total linear momentum $\vec{P}$?
The total momentum of a system of masses (i.e. moving bodies) in any one direction remains constant, unless acted upon by an external force in that direction. This statement is called-
