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.

Reflect on the scenario of a ball sliding down a frictionless slope. Initially at rest, the ball descends through a height h. Applying the principle of conservation of mechanical energy, we deduce that the initial speed $ u_0 $ is given by $ u_0 = \sqrt{2gh} $. Consequently, the velocity of the ball as it reaches the end of the slide can be expressed as :

$ \vec{u}_0 = \sqrt{2gh} \, \hat{\imath} . $

In this scenario where a ball is launched horizontally with an initial speed $ u_0 $ from the top of a building that is $ 3h $ tall, we calculate the time $ t $ it takes for the ball to descend the vertical distance of $ 3h $. This time duration can be determined using the equation of motion under gravity, leading to :

$ t = \sqrt{\frac{2(3h)}{g}} = \sqrt{\frac{6h}{g}}. $

Considering the horizontal distance covered over the time period t, we can express it as :

$ d = u_0 t = \sqrt{2gh} \cdot \sqrt{\frac{6h}{g}} = 2\sqrt{3}h $ ..............(1)

When the ball impacts the ground, its velocity $\vec{u}$ comprises two components : a horizontal component $u_x$ along $\hat{\imath}$ and a vertical component $u_y$ along $\hat{\jmath}$. The horizontal velocity component remains unchanged throughout the motion and is given by $u_x = \sqrt{2gh}$. On the other hand, the vertical velocity component just before the ball hits the ground can be described as :

$ u_y = -\sqrt{2g(3h)} = -\sqrt{6gh} $

Now, let's focus on the ball's impact with the ground. Since the ground can only exert an impulse in the vertical direction, the horizontal momentum of the ball remains unchanged. This means the horizontal component of the ball's velocity after collision (represented as $ v \cos \phi )$ is equal to its horizontal velocity component before the collision. Therefore, we can write this relationship as :

$ v \cos \phi = u_x = \sqrt{2gh} $ ..........(2)

In the vertical axis, the approach velocity, denoted as $ v_a $, equals the absolute value of $ u_y $, which is $ \sqrt{6gh} $. The separation velocity, denoted as $ v_s $, is expressed as $ v \sin \phi $. Using the coefficient of restitution $ e $, which is defined as $ \frac{1}{\sqrt{3}} $, we can establish the relationship :

$ e = \frac{1}{\sqrt{3}} = \frac{v_s}{v_a} = \frac{v \sin \phi}{\sqrt{6gh}}. $

This equation leads to the conclusion that :

$ v \sin \phi = \sqrt{2gh} $ ...............(3)

From these two equations (2) and (3), it's clear that $ v \cos \phi $ and $ v \sin \phi $ are equal. This equality suggests that $ \phi = 45^\circ $, as the sine and cosine of 45 degrees are equal.

To find $ v $, we can use either of the two equations. Let's use the first one :

$ v \cos 45^\circ = \sqrt{2gh} $

Since $ \cos 45^\circ = \frac{1}{\sqrt{2}} $, we can rearrange and solve for $ v $:

$ v = \sqrt{2gh} \times \sqrt{2} = \sqrt{4gh} $

Thus, $ \phi = 45^\circ $ and $ v = \sqrt{4gh} $.

The ball bounces off with velocity $ \vec{v} $ composed of both horizontal and vertical components, as we have established that both $ v \cos \phi $ and $ v \sin \phi $) are equal to $ \sqrt{2gh} $, and the angle $ \phi $ is $ 45^\circ $.

In vector form, the velocity of the ball after bouncing can be represented as the sum of its horizontal and vertical components. Given that both components are equal to $ \sqrt{2gh} $ and considering the direction of these components along the unit vectors $ \hat{\imath} $ (horizontal) and $ \hat{\jmath} $ (vertical), the velocity vector $ \vec{v} $ can be expressed as :

$ \vec{v} = \sqrt{2gh}\hat{\imath} + \sqrt{2gh}\hat{\jmath}. $

$ \Rightarrow $ $ \vec{v} = \sqrt{2gh}(\hat{\imath} + \hat{\jmath}). $

This equation indicates that after the collision, the ball bounces off with equal horizontal and vertical velocity components, each of magnitude $ \sqrt{2gh} $.

The peak height, $ h_1 $, achieved during the projectile's trajectory from point C to D, can be calculated using the formula :

$ h_1 = \frac{v^2 \sin^2 \phi}{2g}. $

In this scenario, where $ v = \sqrt{4gh} $ and $ \phi = 45^\circ $, the expression becomes :

$ h_1 = \frac{(\sqrt{4gh})^2 \sin^2 45^\circ}{2g}. $

Using the fact that $ \sin 45^\circ $ is $ \frac{1}{\sqrt{2}} $ and simplifying, we find that :

$ h_1 = h = \frac{d}{2\sqrt{3}}, $

where the equation (1) has been applied to establish the relationship.

Free body diagram of rod


$\sum {{F _x} = 0} $

${R_1} = T\cos 45^\circ $

${R_1} = {T \over {\sqrt 2 }}$ ..... (i)

$\sum {{F_y} = 0} $

${R_2} + T\sin 45^\circ = W + \alpha W$

${R_2} + {T \over {\sqrt 2 }} = W(1 + \alpha )$ ..... (ii)

$\sum {{\tau _0} = 0} $

$W{L \over 2} + \alpha WL = {T \over {\sqrt 2 }}L$

$T = \sqrt 2 W\left[ {\alpha + {1 \over 2}} \right]$ ..... (iii)

From Eqs. (ii) and (iii), we get

${R_2} + W\left[ {\alpha + {1 \over 2}} \right] = W(1 + \alpha )$

${R_2} = {W \over 2}$

Hence, option (a) is correct.

From Eqs. (i) and (iii), we get

${R_1} = W\left[ {\alpha + {1 \over 2}} \right]$

$\alpha$ = 0.5, R₁ = W

Hence, option (b) is correct.

From Eq. (iii), if $\alpha$ = 0.5

$T = \sqrt 2 W$

${T_{\max }} = 2\sqrt 2 W$

For rope to break,

$T > 2\sqrt 2 W$

$\sqrt 2 W\left[ {\alpha + {1 \over 2}} \right] > 2\sqrt 2 W \Rightarrow \alpha > {3 \over 2}$

Hence, option (d) is correct.

Force acting along $\mathrm{X}$-axis $=m a=0.2 \times 10$

$ =2 \mathrm{~N} $

Distance covered in $2 \mathrm{~s}$ along $\mathrm{X}$-axis

$ =\frac{1}{2} a t^2=\frac{1}{2} \times 10 \times 4=20 \mathrm{~m}=2 l $

Particle reaches at the point $(l,-h)$ at $t=2 \mathrm{~s}$ (Option (A) is correct)

$\tau$ at $(l,-h)=F h=2 \times 1=2 \mathrm{Nm}$

$\vec{\tau}=2(\hat{k}) \quad$ (Option (B) is correct)

$v(t=2 \mathrm{~s})=10 \times 2=20 \mathrm{~m} / \mathrm{s}$

$|\overrightarrow{\mathrm{L}}|$ at $(l,-h)=\mathrm{m} v h=0.2 \times 20 \times 1=4 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}$

$\overrightarrow{\mathrm{L}}=4(\hat{k}) \quad$ (Option $(\mathrm{C})$ is correct)

$\tau$ at $(0,-h)=F h=2 \times 1=2 \mathrm{Nm}$

$\vec{\tau}=2(\hat{k}) \quad$ (Option (D) is incorrect)

Here we will apply the concept of energy conservation and concept of SHM.

We know that for elastic collision, coefficient of restitution, e = 1

Hence, $e = {v \over {0.5{u_0}}} = 1 \Rightarrow v = 0.5{u_0}$

Therefore, particle rebounds with the same speed after collision.

Now, by energy conservation of the spring-mass system, we can say that the speed of the particle when it returns to its equilibrium position is $u_0$.

By SHM,

$v = {v_{\max }}\cos wt$

$ \Rightarrow 0.5{u_0} = {u_0}\cos \left( {\sqrt {{K \over m}} t} \right)$

$ \Rightarrow {1 \over 2} = \cos \left( {\sqrt {{K \over m}} t} \right)$

$ \Rightarrow {\pi \over 3} = \sqrt {{K \over m}} t$

$ \Rightarrow t = {\pi \over 3}\sqrt {{m \over K}} $

Hence, time taken by the particle to reach the wall is ${\pi \over 3}\sqrt {{m \over K}} $

So the time at which the particle passes through the equilibrium position for the first time is

$2t = {{2\pi } \over 3}\sqrt {{m \over K}} $

The time at which the maximum compression of the spring occurs is

$t' = 2t + {T \over 4} = {{2\pi } \over 3}\sqrt {{m \over K}} + {{2\pi } \over 4}\sqrt {{m \over K}} $

$ = \left( {{2 \over 3} + {1 \over 2}} \right)\pi \sqrt {{m \over K}} $

$t' = {{7\pi } \over 6}\sqrt {{m \over K}} $

The time at which the particle passes through the equilibrium position for the second time is

$ = t' + T$

$ = {{7\pi } \over 6}\sqrt {{m \over K}} + {\pi \over 2}\sqrt {{m \over K}} $

$ = \left( {{{7\pi } \over 6} + {\pi \over 2}} \right)\sqrt {{m \over K}} = {{10\pi } \over 6}\sqrt {{m \over K}} $

$ = {{5\pi } \over 3}\sqrt {{m \over K}} $

Hence, options (A) and (D) are correct.

Here, m₁ = 1 kg, m₂ = 5 kg

u₁ = u, u₂ = 0

v₁ = $-$2 m s^$-$1, v₂ = v

By the law of conservation of linear momentum, we get

${m_1}{u_1} + {m_2}{u_2} = {m_1}{u_1} + {m_2}{u_2}$

$1 \times u + 5 \times 0 = 1 \times ( - 2) + 5 \times v$

$u = 5v - 2$ ...... (i)

By the definition of coefficient of restitution,

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

For a perfectly elastic collision, e = 1

$\therefore$ $1 = {{v + 2} \over u}$ or $u = v + 2$ ..... (ii)

Solving equations (i) and (ii), we get

u = 3 m s^$-$1, v = 1 m s^$-$1

Before collision,

Total momentum of the system = 1 $\times$ 3 + 5 $\times$ 0 = 3 kg m s^$-$1

After collision,

Total momentum of the system = 1 $\times$ ($-$2) + 5 $\times$ (1) = 3 kg m s^$-$1

Hence, option (a) is correct.

Momentum of 5 kg mass after collision = 5 $\times$ 1 = 5 kg m s^$-$1.

Hence, option (b) is incorrect.

Velocity of centre mass is ${v_{cm}} = {{1 \times 3 + 5 \times 0} \over {1 + 5}} = {1 \over 2}$ m s ^$-$1

Kinetic energy of the centre of mass

$ = {1 \over 2}{m_{system}}v_{CM}^2 = {1 \over 2} \times 6 \times {\left( {{1 \over 2}} \right)^2} = 0.75$ J

Hence, option (c) is correct.

Before collision,

Total kinetic energy of the system

$ = {1 \over 2} \times 1 \times {3^2} + {1 \over 2} \times 5 \times {0^2} = 4.5$ J

After collision,

Total kinetic energy of the system

$ = {1 \over 2} \times 1 \times {( - 2)^2} + {1 \over 2} \times 5 \times {(1)^2} = 4.5$ J

Hence, option (d) is incorrect.

In free space, there is no external force. Hence linear momentum of the system is conserved. Initial and final linear momentum of the system are

$ \begin{aligned} & \vec{p}_i=\vec{p}_1+\vec{p}_2=\hat{p}-\hat{\imath} \hat{\imath}=\overrightarrow{0}, \\\\ & \vec{p}_f=\vec{p}_1^{\prime}+\vec{p}_2^{\prime} . \end{aligned} $

The conservation of linear momentum, $\vec{p}_i=\vec{p}_f$ gives

$ \vec{p}_1^{\prime}+\vec{p}_2^{\prime}=\overrightarrow{0} . $

In case $(A)$,

$ \begin{aligned} \vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+\left(b_1+b_2\right) \hat{\jmath}+c_1 \hat{k} \\\\ & \neq \overrightarrow{0}, \quad\left(\because c_1 \neq 0\right) . \end{aligned} $

In case $(B)$,

$ \begin{aligned} \vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(c_1+c_2\right) \hat{k} \\\\ & =\overrightarrow{0}, \quad\left(\text { if } c_2=-c_1\right) . \end{aligned} $

In case $(\mathrm{C})$,

$ \begin{aligned} \vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+\left(b_1+b_2\right) \hat{\jmath} \\\\ & =\overrightarrow{0}, \quad\left(\text { if } a_2=-a_1 \text { and } b_2=-b_1\right) . \end{aligned} $

In case (D),

$ \begin{aligned} \vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+2 b_1 \hat{\jmath} \\\\ & \neq \overrightarrow{0}, \quad\left(\because b_1 \neq 0\right) . \end{aligned} $