Impulse & Momentum

$\eqalign{ & By\,\,applying\,\,momentum\,\,conservation\, \cr & along\,\,x\,\,axis\,\,{P_i}\,\,\, = P_f \cr & 2m{v_1}\,\, = \,2m{v_{if}}\cos \theta \, + \,m{v_{2f}}\cos \phi \cr & 2{v_1}\,\,\,\,\, = \,2{v_{if}}\cos \theta \, + {v_{2f}}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \cr & Along\,y - axis,\,\,\,{P_i}\,\,\, = P_f \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = 2m{v_{if\,}}\sin \theta - m{v_{2f}}\sin \phi \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 2{v_{1f}}\sin \theta \,\, = {v_{2f}}\sin \phi \,\,\,\,\,\,\,\,\,\,\,\,...(ii) \cr} $ $By\,\,conservation\,\,of\,\,kinetic\,\,energy,\,\,K{E_i} = K{E_f}$ ${1 \over 2}(2m)v_1^2 = {1 \over 2}(2m)v_{1f}^2 + {1 \over 2}mv_{2f}^2$ $\eqalign{ & v_1^2 = v_{1f}^2 + {1 \over 2}v_{2f}^2 \cr & \Rightarrow 2v_1^2 = 2v_{1f}^2 + v_{2f}^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \cr & from(i),\,\,{(2{v_1} - 2{v_{1f}}\cos \theta )^2} = {({v_{2f}}\cos \phi )^2} \cr & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2{\cos ^2}\theta - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2{\cos ^2}\phi \cr & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2{\cos ^2}\theta - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2(1 - si{n^2}\phi ) \cr & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2{\cos ^2}\theta - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2 - v_{2f}^2{\sin ^2}\phi \cr & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2{\cos ^2}\theta - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2 - 4v_{1f}^2{\sin ^2}\theta (from\,2) \cr} $$\eqalign{ & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2 - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2 \cr & \Rightarrow 4v_{_1}^2 + 4v_{1f}^2 - 8{v_1}{v_{1f}}\cos \theta = v_{2f}^2 - 2v_{1f}^2 \cr & \Rightarrow 6v_{_1f}^2 - 8{v_1}\cos \theta .v_{1f}^{} + 2v_1^2 = 0 \cr & \Rightarrow 3v_{_1f}^2 - 4{v_1}\cos \theta .v_{1f}^{} + v_1^2 = 0 \cr & (This\,\,is\,\,a\,\,quadratic\,\,eqn\,\,in\,\,{v_{if}}) \cr & So\,for\,\,real\,\,roots,\,\,{b^2} - 4ac \ge 0 \cr & \Rightarrow {( - 4{v_1}\cos \theta \,)^2} - 4(3)({v_1}^2) \ge 0 \cr} $$\eqalign{ & {\cos ^2}\theta \ge {3 \over 4} \cr & \Rightarrow \cos \theta \ge {{\sqrt 3 } \over 2}\,\,\,So\,\,\theta \le \,{\pi \over 6} \cr & hence,\,\,\theta = {\pi \over 6} \cr} $

Let's solve the problem step by step. We are given a block of mass $5 \mathrm{~kg}$ subjected to a force $F = (-20x + 10) \mathrm{N}$. To find the position and momentum of the block at $t = \frac{\pi}{4}$ seconds, we'll use the equations of motion and integrate the force to find the acceleration, velocity, and position.

First, we start with Newton's second law:

$F = m \cdot a$

Given:

$m = 5 \mathrm{~kg}$ $F = -20x + 10$

So, the acceleration $a$ can be expressed as:

$a = \frac{F}{m} = \frac{-20x + 10}{5} = -4x + 2$

Acceleration is the second derivative of position with respect to time:

$a = \frac{d^2 x}{dt^2} = -4x + 2$

To solve this differential equation, let's solve it step by step. Assume a solution of the form $x(t) = A \cos(\omega t) + B \sin(\omega t) + C$. Consider using the method of undetermined coefficients:

We first need to find the homogeneous solution (no forcing term):

$\frac{d^2 x_h}{dt^2} + 4x_h = 0$

This is a second-order homogeneous differential equation with constant coefficients. Its characteristic equation is:

$r^2 + 4 = 0$

Solving for $r$:

$r = \pm 2i$

Thus, the general solution to the homogeneous equation is:

$x_h(t) = A \cos(2t) + B \sin(2t)$

Next, we need a particular solution to the non-homogeneous equation:

$x_p = C$

Substituting $x_p$ into the differential equation:

$a = -4C + 2 = 0 \implies C = \frac{1}{2}$

Therefore, the general solution of the differential equation is:

$x(t) = A \cos(2t) + B \sin(2t) + \frac{1}{2}$

Using the initial conditions to determine $A$ and $B$:

At $t = 0$, $x(0) = 1$:

$1 = A \cdot \cos(0) + B \cdot \sin(0) + \frac{1}{2} \implies 1 = A + \frac{1}{2} \implies A = \frac{1}{2}$

Since the block is initially at rest, $\frac{dx}{dt}|_{t=0} = 0$:

$\frac{dx}{dt} = -2A \sin(2t) + 2B \cos(2t)$

At $t = 0$:

$0 = -2A \cdot \sin(0) + 2B \cdot \cos(0) = 2B \implies B = 0$

Therefore, the position equation simplifies to:

$x(t) = \frac{1}{2} \cos(2t) + \frac{1}{2}$

Now, we find the position at $t = \frac{\pi}{4}$ seconds:

$x\left(\frac{\pi}{4}\right) = \frac{1}{2} \cos\left(2 \cdot \frac{\pi}{4}\right) + \frac{1}{2} = \frac{1}{2} \cos\left(\frac{\pi}{2}\right) + \frac{1}{2} = \frac{1}{2} \cdot 0 + \frac{1}{2} = \frac{1}{2} \mathrm{~m}$

To find the momentum, we first find the velocity by differentiating the position function:

$v(t) = \frac{dx}{dt} = -\sin(2t)$

Thus, the velocity at $t = \frac{\pi}{4}$ is:

$v\left(\frac{\pi}{4}\right) = -\sin\left(2 \cdot \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$

Finally, the momentum $p$ is given by:

$p = m \cdot v = 5 \mathrm{~kg} \cdot -1 \mathrm{~m/s} = -5 \mathrm{~kg \cdot m/s}$

Therefore, the position and momentum of the block at $t = \frac{\pi}{4} \mathrm{s}$ are:

Position: $0.5 \mathrm{~m}$

Momentum: $-5 \mathrm{~kg \cdot m/s}$

The correct option is:

Option C: $0.5 \mathrm{~m}, -5 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$

The situation is depicted in the figure below.

At the highest point of the first particle’s trajectory, its speed is given by :

$ \text{Speed of the first particle} = u_0 \cos \alpha $

For the second particle, which was thrown vertically upward, the speed at its highest point is calculated as :

$ \text{Speed of the second particle} = \sqrt{u_0^2 - 2gH} $

Where $ H $ is the maximum height reached by the second particle :

$ H = \frac{u_0^2 \sin^2 \alpha}{2g} $

Therefore, the speed of the second particle at the highest point becomes :

$ = \sqrt{u_0^2 - \frac{2g \left( u_0^2 \sin^2 \alpha \right)}{2g}} = \sqrt{u_0^2 - u_0^2 \sin^2 \alpha} = u_0 \cos \alpha $

Thus, at the highest point :

According to the law of conservation of linear momentum in the horizontal direction :

$ mu_0 \cos \alpha = 2mv_1 $

$ \implies v_1 = \frac{u_0 \cos \alpha}{2} \quad \text{... (i)} $

In the vertical direction :

$ mu_0 \cos \alpha = 2mv_2 $

$ \implies v_2 = \frac{u_0 \cos \alpha}{2} \quad \text{... (ii)} $

Let $ \theta $ be the angle that the composite system makes with the horizontal immediately after the collision. Then :

$ \tan \theta = \frac{v_2}{v_1} = 1 \quad \text{(Using (i) and (ii))} $

$ \therefore \theta = \tan^{-1} 1 = 45^\circ = \frac{\pi}{4} $

The time of flight for the bullet is same as that of the ball and is given by

$t = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 5} \over {10}}} = 1\,s$

Just after the collision, the velocity of bullet (v₁) is related to its range (R₁) by v₁t = R₁ which gives v₁ = 100 m/s. Similarly, the velocity of the ball is v₂ = 20 m/s. Consider the bullet and the ball together as a system. Along the direction of collision, there is no external force on the system. Hence, linear momentum of the system in the direction of collision is conserved. The linear momentum of the system before and after the collision are

${p_i} = {m_1}V$,

${p_f} = {m_1}{v_1} + {m_2}{v_2}$.

The conservation of linear momentum, ${p_i} = {p_f}$, gives

$V = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1}}}$

$ = {{(0.01)(100) + (0.2)(20)} \over {0.01}} = 500$ m/s.

The coordinates of centre of mass is defined as

${R_{CM}} = {{\sum\nolimits_i {{m_i}{r_i}} } \over {\sum\nolimits_i {{m_i}} }}$

Thus,

${Y_{CM}} = {{(6m \times 0) + (m \times a) + (m \times a) + (m \times 0) + (m \times - a)} \over {6m + m + m + m + m}} = {a \over {10}}$

The faster particle covers twice the distance covered by the slower one and they meet each other whenever the slower one covers an angular displacement of $2\pi/3$. since the masses are same at each collision, they interchange the velocity. The number of meeting points is $\frac{2\pi}{2\pi/3}=3$ points; hence, they meet at two points other than A.

${v_V} = v\sin 30^\circ \cos 30^\circ - v\cos 30^\circ \cos 60^\circ = 0$

Statement 1 : For an elastic collision, the coefficient of restitution = 1

$e = \left| {{{{v_2} - {v_1}} \over {{u_1} - {u_2}}}} \right|$

$ \Rightarrow |{v_2} - {v_1}| = |{u_1} - {u_2}|$

Relative velocity after collision is equal to relative velocity before collision. But in the statement relative speed is given.

Statement 2 : Linear momentum remains conserved in an elastic collision. This statement is true.

Hint :

Relative velocity before collision $ = |{u_1} - {u_2}|$

Relative velocity after collision $ = |{v_1} - {v_2}|$

Elastic collision $ = e = 1$

$e = \left| {{{{v_1} - {v_2}} \over {{u_1} - {u_2}}}} \right| = 1$

$ \Rightarrow |{v_1} - {v_2}| = |{u_1} - {u_2}|$

So statement 1 false

$\sum {{F_{ext}} = 0} $, So momentum should be conserved.

Statement 2 is true.