Center of Mass and Collision

Given:

Mass of the wedge (M) = 4m

Mass of the particle (m)

Initial speed of the particle (v)

There is no friction.

Step 1 : Conservation of Linear Momentum

Before the collision, the momentum of the system is just the momentum of the particle because the wedge is at rest. After the collision, both the particle and the wedge will be moving. Let's denote the final common velocity as $v'$.

We can write the conservation of momentum as :

$mv = (m + 4m)v'$

which simplifies to

$v' = \frac{v}{5} \tag{1}$

Step 2 : Conservation of Mechanical Energy

We apply the conservation of energy before and after the collision. Before the collision, only the particle has kinetic energy. After the collision, both the particle and the wedge have kinetic energy and the particle has potential energy due to its height h on the wedge.

We can write the conservation of energy as :

$\frac{1}{2}mv^2 = \frac{1}{2}(m + 4m){v'}^2 + mgh$

which simplifies to

$mv^2 = (m + 4m){v'}^2 + 2mgh$

Substituting equation (1) into this equation gives

$v^2 = 5{v'}^2 + 2gh \Rightarrow \frac{4}{5}v^2 = 2gh$

which simplifies to

$h = \frac{2v^2}{5g}$

So, the maximum height climbed by the particle on the wedge is given by $\frac{2v^2}{5g}$.

Therefore, the correct answer is Option D :

$\frac{2v^2}{5g}$

Momentum imparted to the surface in one collision,

$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$ ..... (i)

Force on the surface due to n collision per second, $F = {n \over t}(\Delta p) = n\Delta p$ ($\because$ $t = 1s$)

= 2 mnv [from Eq. (i)]

So, pressure on the surface,

$p = {F \over A} = {{2mnv} \over A}$

Here, m = 10^$-$26 kg, n = 10²² s^$-$1, v = 10⁴ ms^$-$1, A = 1 m²

$\therefore$ Pressure, $p = {{2 \times {{10}^{ - 26}} \times {{10}^{22}} \times {{10}^4}} \over 1} = 2$ N/m²

So, pressure exerted is of order of 10⁰.

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v₁ is initial velocity.

(i) Before collision :

Now, in figure (ii), v₁ is final velocity of unknown particle and v₂ is final velocity of proton.

(ii) After collision :

By conservation of momentum, we have

Momentum before collision = Momentum after collision

Consider x-component, we have

$m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ $

$m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})$ ..... (1)

Consider y-component, we have

$0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ $

${1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}$ ...... (2)

Substitute Eq. (2) in Eq. (1), we get

$m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}$

$ \Rightarrow {v_i} = \sqrt 2 {v_2}$ ..... (3)

Using Eq. (2) and (3) in Eq. (1), we get

$m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})$

$2m{v_2} = 2m'{v_1}$ ($\because$ ${v_1} = {v_2}$)

$ \Rightarrow m = m'$

Statement ${\rm I}{\rm I}$ says "Maximum energy loss occurs when the particles get stuck together as a result of the collision." This is a case of perfectly inelastic collision.

Hence statement ${\rm I}$${\rm I}$ is correct.

Given The linear mass density $\lambda = k{\left( {{x \over L}} \right)^n}$

The mass of the removed disc $ = \sigma \left( {\pi {R^2}} \right) = \pi \sigma {R^2}$

So mass of the remaining disc = $4\pi \sigma {R^2}$ - $\pi \sigma {R^2}$ = $3\pi \sigma {R^2}$

Let center of mass of $3\pi \sigma {R^2}$ mass is at x distance from origin O.

$\therefore$ ${{3\pi {R^2}\sigma .x + \pi {R^2}\sigma .R} \over {4\pi {R^2}\sigma }} = 0$

As center of mass of full disc is at Origin.

$\therefore$ $x = - {R \over 3}$

According to the question, $x$ = $\alpha R$

$\therefore$ $\alpha = - {1 \over 3}$

$ \Rightarrow $ $\left| \alpha \right| = {1 \over 3}$

Finally,

mass m₁ moved towards the center a distance d so the distance of mass m₁ from the origin is x₁ - d, and now let mass m₂ need to move d' to keep the center at the origin.

$\therefore$ $0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}$

$ \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$

$ \Rightarrow d' = {{{m_1}} \over {{m_2}}}d$

In $y$-direction, ${{mv} \over {\sqrt 3 }} = m{v_1}\,\sin \theta \,\,\,\,\,\,\,\,\,\,...(2)$

Squaring and adding eqns.$(1)$ and $(2)$

${\left( {m{v_1}\cos \theta } \right)^2} + {\left( {m{v_1}\sin \theta } \right)^2}$$ = {\left( {mv} \right)^2} + {\left( {{{mv} \over {\sqrt 3 }}} \right)^2}$

$ \Rightarrow $ $v_1^2 = {v^2} + {{{v^2}} \over {\sqrt 3 }} $

$\Rightarrow {v_1} = {2 \over {\sqrt 3 }}v$

The correct answer is Option A : $A$ does not imply $B$ and $B$ does not imply $A$.

Here's why :

Statement $A:$ The linear momentum of a system of particles being zero does not mean that the kinetic energy is also zero. For example, consider two equal mass particles moving with the same speed but in opposite directions. The linear momentum of the system will be zero because momentum is a vector quantity and the two momenta will cancel out. However, kinetic energy is a scalar quantity and does not cancel out in this way. Each particle has kinetic energy due to its motion, so the total kinetic energy of the system is not zero.

Statement $B:$ The kinetic energy of a system of particles being zero also does not imply that the linear momentum is zero. If the kinetic energy is zero, it means that all the particles are at rest (since kinetic energy is associated with motion). However, the linear momentum will also be zero in this case because momentum depends on both mass and velocity, and the velocity of each particle is zero.

But remember that zero linear momentum doesn't exclusively mean all particles are at rest. As explained above, it could be the scenario where particles have equal and opposite momenta, thereby cancelling each other. Hence, $B$ doesn't imply $A$.

So, neither statement implies the other. Hence, Option A is correct.