mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

1. a) A particle of mass $m$$m$mm$m$ moving to the right with an initial velocity $u$$u$uu$u$ collides elastically with a particle of unknown mass $M$$M$MM$M$ coming from the opposite direction. After the collision $m$$m$mm$m$ has a velocity $u/2$$u/2$u//2u / 2$u/2$ at right angles to the incident direction, and $M$$M$MM$M$ is deflected back making an angle of ${45}^{\circ }$${45}^{\circ }$45^(@)45^{\circ}${45}^{\circ }$ degrees to its incident direction as shown below. Calculate the ratio $M/m$$M/m$M//mM / m$M/m$.

1. Momentum Conservation (Horizontal Direction):
   Before the collision, the total horizontal momentum is $mu-Mu$$mu-Mu$mu-Mumu – Mu$mu-Mu$. After the collision, the horizontal momentum of $m$$m$mm$m$ is zero (since it moves at right angles to the incident direction), and the horizontal component of $M$$M$MM$M$‘s momentum is $Mv\cos ({45}^{\circ })$$Mv\cos ({45}^{\circ })$Mv cos(45^(@))Mv \cos(45^\circ)$Mv\cos ({45}^{\circ })$, where $v$$v$vv$v$ is the final velocity of $M$$M$MM$M$. Therefore, we have:
   $$mu-Mu=Mv\cos ({45}^{\circ })$$$$mu-Mu=Mv\cos ({45}^{\circ })$$mu-Mu=Mv cos(45^(@))mu – Mu = Mv \cos(45^\circ)$$mu-Mu=Mv\cos ({45}^{\circ })$$
   Since $\cos ({45}^{\circ })=\frac{1}{\sqrt{2}}$$\cos ({45}^{\circ })=\frac{1}{\sqrt{2}}$cos(45^(@))=(1)/(sqrt2)\cos(45^\circ) = \frac{1}{\sqrt{2}}$\cos ({45}^{\circ })=\frac{1}{\sqrt{2}}$, this equation simplifies to:
   $$mu-Mu=\frac{Mv}{\sqrt{2}}$$$$mu-Mu=\frac{Mv}{\sqrt{2}}$$mu-Mu=(Mv)/(sqrt2)mu – Mu = \frac{Mv}{\sqrt{2}}$$mu-Mu=\frac{Mv}{\sqrt{2}}$$
   $$u(m-M)=\frac{Mv}{\sqrt{2}}\text{(1)}$$$$u(m-M)=\frac{Mv}{\sqrt{2}}\text{(1)}$$u(m-M)=(Mv)/(sqrt2)quad(1)u(m – M) = \frac{Mv}{\sqrt{2}} \quad \text{(1)}$$u(m-M)=\frac{Mv}{\sqrt{2}}\text{(1)}$$
2. Momentum Conservation (Vertical Direction):
   Before the collision, the total vertical momentum is zero. After the collision, the vertical momentum of $m$$m$mm$m$ is $mu/2$$mu/2$mu//2mu/2$mu/2$, and the vertical component of $M$$M$MM$M$‘s momentum is $Mv\sin ({45}^{\circ })$$Mv\sin ({45}^{\circ })$Mv sin(45^(@))Mv \sin(45^\circ)$Mv\sin ({45}^{\circ })$. Therefore:
   $$0=mu/2-Mv\sin ({45}^{\circ })$$$$0=mu/2-Mv\sin ({45}^{\circ })$$0=mu//2-Mv sin(45^(@))0 = mu/2 – Mv \sin(45^\circ)$$0=mu/2-Mv\sin ({45}^{\circ })$$
   Using $\sin ({45}^{\circ })=\frac{1}{\sqrt{2}}$$\sin ({45}^{\circ })=\frac{1}{\sqrt{2}}$sin(45^(@))=(1)/(sqrt2)\sin(45^\circ) = \frac{1}{\sqrt{2}}$\sin ({45}^{\circ })=\frac{1}{\sqrt{2}}$, this equation simplifies to:
   $$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$$$$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$$(mu)/(2)=(Mv)/(sqrt2)\frac{mu}{2} = \frac{Mv}{\sqrt{2}}$$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$$
   $$mu=Mv\sqrt{2}\text{(2)}$$$$mu=Mv\sqrt{2}\text{(2)}$$mu=Mvsqrt2quad(2)mu = Mv\sqrt{2} \quad \text{(2)}$$mu=Mv\sqrt{2}\text{(2)}$$
3. Kinetic Energy Conservation:
   The total kinetic energy before and after the collision remains the same. Before the collision, the total kinetic energy is $\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2$$\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2$(1)/(2)mu^(2)+(1)/(2)Mu^(2)\frac{1}{2}mu^2 + \frac{1}{2}Mu^2$\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2$. After the collision, it is $\frac{1}{2}m(u/2{)}^2+\frac{1}{2}M{v}^2$$\frac{1}{2}m(u/2{)}^2+\frac{1}{2}M{v}^2$(1)/(2)m(u//2)^(2)+(1)/(2)Mv^(2)\frac{1}{2}m(u/2)^2 + \frac{1}{2}Mv^2$\frac{1}{2}m(u/2{)}^2+\frac{1}{2}M{v}^2$. Equating these, we get:
   $$\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2+\frac{1}{2}M{v}^2$$$$\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2+\frac{1}{2}M{v}^2$$(1)/(2)mu^(2)+(1)/(2)Mu^(2)=(1)/(2)m((u)/(2))^(2)+(1)/(2)Mv^(2)\frac{1}{2}mu^2 + \frac{1}{2}Mu^2 = \frac{1}{2}m\left(\frac{u}{2}\right)^2 + \frac{1}{2}Mv^2$$\frac{1}{2}m{u}^2+\frac{1}{2}M{u}^2=\frac{1}{2}m{\left(\frac{u}{2}\right)}^2+\frac{1}{2}M{v}^2$$
   Simplifying, we find:
   $$m{u}^2+M{u}^2=\frac{1}{4}m{u}^2+M{v}^2$$$$m{u}^2+M{u}^2=\frac{1}{4}m{u}^2+M{v}^2$$mu^(2)+Mu^(2)=(1)/(4)mu^(2)+Mv^(2)mu^2 + Mu^2 = \frac{1}{4}mu^2 + Mv^2$$m{u}^2+M{u}^2=\frac{1}{4}m{u}^2+M{v}^2$$
   $$\frac{3}{4}m{u}^2+M{u}^2=M{v}^2\text{(3)}$$$$\frac{3}{4}m{u}^2+M{u}^2=M{v}^2\text{(3)}$$(3)/(4)mu^(2)+Mu^(2)=Mv^(2)quad(3)\frac{3}{4}mu^2 + Mu^2 = Mv^2 \quad \text{(3)}$$\frac{3}{4}m{u}^2+M{u}^2=M{v}^2\text{(3)}$$