Center of Mass , Momentum and Collision - NEET Physics Chapterwise MCQs & PYQs

NEET Center of Mass , Momentum and Collision MCQs & PYQs

Practice NEET Center of Mass , Momentum and Collision Questions

Question 61:

difficult

Two particles of masses $m_1$, $m_2$ move with initial velocities $u_1$ and $u_2$. On collision, one of the particles get excited to higher level, after absorbing energy $\varepsilon$. If final velocities of particles be $v_1$ and $v_2$, then we must have:

(2015)

According to the conservation of energy, the total initial energy equals the total final energy plus the energy absorbed ($\varepsilon$). Thus, initial kinetic energy minus absorbed energy equals final kinetic energy: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \varepsilon = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$.

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Question 62:

moderate

On a frictionless surface, a block of mass $M$ moving at speed $v$ collides elastically with another block of same mass $M$ which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $v/3$. The second block’s speed after the collision is:

(2015 Re)

In an elastic collision between two equal masses where one is initially at rest, the angle between final velocities is $90^\circ$, yielding $v^2 = v_1^2 + v_2^2$. Substituting $v_1 = v/3$, we get $v_2 = \sqrt{v^2 - (v/3)^2} = \frac{2\sqrt{2}}{3}v$.

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Question 63:

moderate

A ball is thrown vertically downwards from a height of $20\text{ m}$ with an initial velocity $u_0$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $u_0$ is: (Take $g = 10\text{ ms}^{-2}$)

(2015 Re)

The velocity just before impact is $v^2 = u_0^2 + 2gh$. Since it loses $50\%$ energy and reaches the same height $h$, the post-collision kinetic energy satisfies $mgh = \frac{1}{2}(\frac{1}{2}mv^2)$, leading to $v^2 = 4gh$. Solving gives $u_0 = \sqrt{2gh} = 20\text{ m/s}$.

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Question 64:

moderate

Two particles $A$ and $B$, move with constant motion in one dimensional with velocities $\vec{v}_1$ and $\vec{v}_2$. At the initial moment their position vectors are $\vec{r}_1$ and $\vec{r}_2$ respectively. The condition for particle $A$ and $B$ for their collision is:

(2015 Re)

For two particles to collide, their relative position vector must be parallel to their relative velocity vector. Hence, the unit vector of relative position must equal the unit vector of relative velocity: $\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$.

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Question 65:

moderate

Two spheres $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. A is at rest initially and B is moving with velocity $v$ along x-axis. After collision B has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

(2012 Pre)

Using conservation of linear momentum along y-axis, $m_1 v_{1y} = -m_2 (v/2)$. Along x-axis, $m_1 v_{1x} = m_2 v$. The angle with the x-axis is given by $\theta = \tan^{-1}(v_{1y}/v_{1x}) = \tan^{-1}(-1/2)$.

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Question 66:

easy

A mass $m$ moving horizontally (along the $x$-axis) with velocity $v$ collides and sticks to a mass of $3\text{ m}$ moving vertically upward (along the $y$-axis) with velocity $2v$. The final velocity of the combination is:

(2011 Mains)

By conservation of momentum, total initial momentum vector is $\vec{P} = mv\hat{i} + (3m)(2v)\hat{j}$. Dividing by the total mass $4m$ gives the final velocity vector $\vec{v}_f = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$.

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Question 67:

easy

Two objects of mass $10\text{ kg}$ and $20\text{ kg}$ respectively are connected to the two ends of a rigid rod of length $10\text{ m}$ with negligible mass. The distance of the centre of mass of the system from the $10\text{ kg}$ mass is :

(2022)

Centre of mass formula is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Taking $10\text{ kg}$ at origin and $20\text{ kg}$ at $10\text{ m}$, we get $x_{cm} = \frac{10(0) + 20(10)}{10+20} = \frac{20}{3}\text{ m}$. Option (c) is correct.

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Question 68:

easy

Two particles of mass $5\text{ kg}$ and $10\text{ kg}$ respectively are attached to the two ends of a rigid rod of length $1\text{ m}$ with negligible mass. The centre of mass of the system from the $5\text{ kg}$ particle is nearly at a distance of :

(2020)

Use the centre of mass formula $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Substituting $m_1 = 5\text{ kg}$, $x_1 = 0$, $m_2 = 10\text{ kg}$, $x_2 = 100\text{ cm}$, we get $x_{cm} = \frac{10 \times 100}{15} = 66.67\text{ cm} \approx 67\text{ cm}$. Option (b) is correct.

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Question 69:

easy

Which of the following statements are correct?

A. Centre of mass of a body always coincides with the centre of gravity of the body

B. Centre of gravity of a body is the point at which the total gravitational torque on the body is zero

C. A couple on a body produce both translational and rotational motion in a body

D. Mechanical advantage greater than one means that small effort can be used to lift a large load

(2017-Delhi)

Centre of gravity is the point where total gravitational torque is zero (Statement B is correct). Mechanical advantage greater than one implies a small effort lifts a large load (Statement D is correct). Thus, statements B and D are correct, making option (d) the right choice.

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Question 70:

moderate

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is :

(2015)

The centre of mass remains stationary. Initial distance of COM from mass $M$ is $\frac{5M \times 12R}{M + 5M} = 10R$. At collision, distance between centers is $3R$, and the distance of $M$ from COM is $\frac{5}{6} \times 3R = 2.5R$. Thus, distance covered by $M$ is $10R - 2.5R = 7.5R$. Option (b) is correct.

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