Center of Mass and Collision

• To reach the horizontal position, velocity of bob should be $=10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

• Velocity of bob and point mass will be interchanged after elastic collision

By conservation of energy

$ \begin{aligned} & m g h=\frac{1}{2} m v^2 \\ & h=\frac{v^2}{2 g} \\ & h=\frac{(10)^2}{2 \times 10} \\ & h=5 \mathrm{~m} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \vec{P}_{\text {net }}=-m v \sin \theta \hat{i}-m v \cos \theta \hat{j}+m v \sin \theta \hat{i}-m v \cos \theta \hat{j} \\ & \vec{P}_{\text {net }}=-2 m v \cos \theta \hat{j} \\ & \text { at } \theta=\frac{\pi}{2}, \vec{P}_{\text {net }}=0, \text { at } \theta=-\frac{\pi}{2}, \vec{P}_{\text {net }}=0 \end{aligned}\\ &\text { So most appropriate graph is } \end{aligned} $

$\begin{aligned} & v_1=\sqrt{2 g h_1} \\ & =\sqrt{2 \times 9.8 \times 40} \\ & v_1=\sqrt{784}=28 \mathrm{~m} \mathrm{~s}^{-1} \\ & \text { and } v_2=\sqrt{2 g h_2}=\sqrt{2 \times 9.8 \times 10} \\ & =\sqrt{196}=14 \mathrm{~m} \mathrm{~s}^{-1} \\ & \text { Impulse }=\Delta \vec{p}=m\left(\vec{v}_f-\vec{v}_i\right)=m\left(\vec{v}_2-\vec{v}_1\right) \\ & =\frac{1}{2}(14-(-28)) \\ & =21 \mathrm{NS} \end{aligned}$

In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body $A$ is moving with a velocity $v_1$ and body $B$ is at rest. The conservation of momentum must hold true because no external forces are acting on the system.

The momentum before the collision is only due to body $A$ since body $B$ is at rest. Therefore, the total initial momentum $p_{\text{initial}}$ of the system is given by:

$p_{\text{initial}} = m_A v_1 + m_B v_0 = mv_1 + 0 = mv_1$

where:

• $m_A$ and $m_B$ are the masses of bodies $A$ and $B$ respectively,


• $m$ is the mass of each body,


• $v_1$ is the velocity of body $A$,


• $v_0 = 0$ is the velocity of body $B$ because it is initially at rest.

Since they undergo a completely inelastic collision, bodies $A$ and $B$ stick together after the collision and hence move with a common velocity $v_2$. The total mass of the combined system post-collision is $m_A + m_B = m + m = 2m$. The momentum after the collision is given by:

$p_{\text{final}} = (m_A + m_B)v_2 = 2m v_2$

Applying the conservation of momentum (since no external force implies momentum is conserved), we equate $p_{\text{initial}}$ and $p_{\text{final}}$:

$mv_1 = 2m v_2$

Dividing through by $m$ yields:

$v_1 = 2v_2$

Thus, solving for the ratio $\frac{v_1}{v_2}$ gives:

$\frac{v_1}{v_2} = 2$

Therefore, the ratio of $v_1$ to $v_2$ is $2:1$, making the correct answer:

Option B: $2:1$

In elastic collision maximum energy is transfer when $M=m$

Final velocity of centre of mass = Initial velocity of centre of mass = 0 because net external force on system is zero.

$F=\left|\frac{\Delta \vec{p}}{\Delta t}\right|=\frac{2 m v \sin \theta}{\mathrm{t}}=\frac{2(1)(1) \sin 60^{\circ}}{0.1}=10 \sqrt{3} \mathrm{~N}$

$P = mv$

$0.3 = 5 \times {10^{ - 3}}\,v$

$v = {{300} \over 2}$

$v = 60$ m/s

$S = v \times t$

$ = 60 \times 5$

$ = 300$ m

As per Newton's II law for a system

${\overrightarrow F _{external}} = {{d\overrightarrow P } \over {dt}} = m{{d{{\overrightarrow v }_{com}}} \over {dt}}$

As explosion occurs due to internal forces only, the velocity of centre of mass would remain unchanged. This means that the centre of mass would continue to travel along same path as it was moving before explosion.

The fragments can however move individually along various directions.

Thus (A) is not correct and (R) is correct.

Momentum of the system would remain conserved.

Initial momentum = 0

Final momentum should also be zero.

Let masses be 2m, 2m and m

Momentum along x-direction = 2mv$\widehat i$

Momentum along y-direction = 2mv$\widehat j$

Net momentum = $\sqrt {{{(2mv)}^2} + {{(2mv)}^2}} = \sqrt 2 \,.\,2$ mv

Now, 2$\sqrt2$mv = mv'

v' = 2$\sqrt2$v

${X_{cm}} = {{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$

$ = {{10 \times 0 + 20 \times 10} \over {10 + 20}}$

$ = {{200} \over {30}}$

$ = {{20} \over 3}$ m

To calculate the impulse imparted to the ball, we need to look at the change in momentum during the collision with the ground. The impulse $ I $ can be determined by the following relationship:

$ I = \Delta p $

Where:

• $ \Delta p $ is the change in momentum.
• Momentum $ p $ is defined as the product of the mass $ m $ of an object and its velocity $ v $.

Since the ball rebounds to the same height, its speed just before it hits the ground and just after it leaves the ground will be the same (ignoring air resistance), although the direction of the velocity will change.

Let's calculate the speed of the ball just before the impact. The ball drops from a height $ h $ with initial velocity $ u = 0 $ m/s. Using the kinematic equation for constant acceleration under gravity $ g $, we have:

$ v^2 = u^2 + 2gh $

Plugging in the values for $ u = 0 $, $ g = 10 $ m/s², and $ h = 10 $ m, we get:

$ v^2 = 0^2 + 2 \cdot 10 \cdot 10 $

$ v^2 = 200 $

$ v = \sqrt{200} $

$ v = 10\sqrt{2} $ m/s

Since the ball bounces back to the same height, its speed on the way up just after the collision will be $ 10\sqrt{2} $ m/s but in the opposite direction.

The change in velocity $ \Delta v $ is:

$ \Delta v = v_{\text{final}} - v_{\text{initial}} $

$ \Delta v = 10\sqrt{2} - (-10\sqrt{2}) $

$ \Delta v = 20\sqrt{2} $ m/s

Given that the mass $ m $ is 0.15 kg, we can now calculate $ \Delta p $:

$ \Delta p = m \Delta v $

$ \Delta p = 0.15 \cdot 20\sqrt{2} $

$ \Delta p = 3\sqrt{2} $

$ \Delta p \approx 3 \cdot 1.414 $

$ \Delta p \approx 4.242 $ kg m/s

The magnitude of the impulse imparted to the ball is therefore approximately 4.242 kg m/s, which most closely matches:

Option C: 4.2 kg m/s.