Mechanics Questions

From Newton's second law of motion, we have, $\frac{d \vec{P}}{dt} = \vec{F} \\$ $m⋅\frac{d \vec{v}}{dt} = \vec{F} \\$ $m⋅\frac{d \vec{v}}{dt} ⋅ \vec{v}= \vec{F} ⋅ \vec{v}\\$ $m⋅\frac{d }{dt} (\frac{v^2}{2})= \vec{F} ⋅ \vec{v}\\$ $\frac{d }{dt} (\frac{mv^2}{2})= \vec{F} ⋅ \vec{v}\\$ $\implies \frac{dT}{dt} = \vec{F} ⋅\vec{v}$

The bomb is exploding, so no external force act on it. Thus the momentum is conserved. $\\$ Mass of the $3^{rd}$ piece is $20 \ kg$ $\\$ Applying conservation of momentum, $\vec{p_i} = \vec{p_f} \\ $ $50 (20 \hat{i}+ 22 \hat{j} +10 \hat{k}) = 10(100 \hat{i}+ 50 \hat{j} + 20 \hat{k}) + 20(30 \hat{i} - 20 \hat{j} - 10 \hat{k}) + 20\vec{v}$ $\\ \implies \vec{v} = (-30 \hat{i} + 50 \hat{j} + 25 \hat{k}) \ ms^{-1}$

In this case, $$ \begin{matrix} E_{before} = m_πc^2, & \bold{p}_{before} = 0, \\ \\ E_{after}= E_μ + E_v, & \bold{p}_{after} = \bold{p}_μ +\bold{p}_v \end{matrix} $$ Conservation of momentum requires that ,$$\bold{p}_v = -\bold{p}_μ $$ Conservation of energy says that, $$E_μ + E_v = m_{π}c^2 \\$$ Now, using \(E=pc\), \(E_v = |\bold{p}_v|c,\) whereas, $$\\$$ using \(E^2 - p^2c^2 = m^2c^4, |\bold{p}_μ| = \frac{\sqrt{{E_μ}^2 - {m_μ}^2c^4 }}{c} \\\) $$\\$$ so, $$E_μ + \sqrt{{E_μ}^2 - {m_μ}^2c^4 } = m_πc^2,$$ $$ \Rightarrow E_μ = \frac{({m_μ}^2+{m_π}^2)c^2}{2m_π} $$

Suppose we transform from $S$ frame to $S'$ frame using Lorentz transformation. So we would replace $x,y,z,t$ with $x',y',z',t'$ using the following relation:- $$ x = γ(x'+vt') \\ y = y' \\ z = z' \\ t = γ(t' + \frac{vx'}{c^2}) $$ where $v$ is the velocity at which $S'$ frame is moving with respect to the $S$ frame along the $X-X'$ frame. $γ = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is relativistic factor. $\\$ so, $$ = [γ(x'+vt')]^2 + (y')^2 + (z')^2 - c^2[γ^2(t' + \frac{vx'}{c^2})^2] \\ = γ^2((x')^2+(vt')^2+2x'vt') + (y')^2 + (z')^2 - c^2γ^2[t'+\frac{vx'}{c^2}]^2 \\ = x'^2+y'^2+z'^2-c^2t'^2 $$ Hence we saw that upon transformation from $S$ to $S'$, one could see there is no change in the nature of the equation $x^2 + y^2 + z^2 - c^2t^2$. Hence the given quantity is invariant under the Lorentz transformation.

The lorentz factor is given by \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \) $$\\$$ here \(v = \frac{3}{5}c \), so \( \gamma = \frac{1}{\sqrt{1-(\frac{3}{5})^2}} = \frac{5}{4}, \\ \) so it lives longer (than at rest) by a factor of \(\frac{5}{4} :\\ \) $$\\$$ \( \frac{5}{4} \times (2 \times 10^{-6})s = 2.5 \times 10^{-6}s \)

Here pion is coming to rest and disintegrates into muon and neutrino with the given rest mass. We would apply:- <ol> <li>conservation of momentum</li> <li>conservation of energy</li> </ol> from the conservation of momentum, we get that magnitude of the momentum of pion and neutrino are the same [$p_π = p_μ + p_n$]. Hence let $p$ be the momentum of both. $\\$ Now, let's apply the conservation of energy:- $$\tag{1} E_π = E_μ + E_n$$ where $E_π$ is the energy of π-meson, $E_μ$ is the energy of muon and $E_n$ is the energy of the neutrino. $$\tag{2} \implies m_πc^2 = (T+ m_μc^2) +pc $$ we used:- $E^2 = p^2c^2 + (m_0c^2)^2$ and $E = T + m_0c^2$, where $T$ is the kinetic energy, also $$\tag{3} m_πc^2 = \sqrt{p^2c^2+ m_μ^2c^4} +pc $$ equation (1) could be written as (2) and (3), from equation (2) we get:- $$\tag{4} T = m_πc^2 - m_μc^2 - pc $$ now we are required to find $pc$ from equation (3) to get $T$ for muon. $\\$ using equation (3) and squaring it:- $$ \begin{matrix} (m_πc^2-pc)^2 = [\sqrt{p^2c^2 + m_μ^2c^4}]^2 \\ \implies m_π^2c^4 + p^2c^2 - 2m_πc^2pc = p^2c^2 + m_μ^2c^4 \\ \implies m_π^2c^4 - m_μ^2c^4 = 2m_πc^2pc \end{matrix}$$ $$ \tag{5} pc = \frac{m_π^2c^4 - m_μ^2c^4}{2m_πc^2} $$ now using equation (5) in equation (4) we get the desired answer. $$T = m_πc^2 - m_μc^2 - \frac{m_π^2c^4 - m_μ^2c^4}{2m_πc^2} $$ which gives us $$ T_μ = \frac{(m_π - m_μ)^2 c^2}{2m_π} $$