Question 41:
easyA ball is dropped from a height of $5\text{ m}$, if it rebound upto height of $1.8\text{ m}$, then the ratio of velocities of the ball after and before rebound is:
(1998)
The velocity before rebound is $v_1 = \sqrt{2gh_1}$ and after rebound is $v_2 = \sqrt{2gh_2}$. The ratio is $\frac{v_2}{v_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{1.8}{5}} = \frac{3}{5}$.
Question 42:
easyA moving body of mass $m$ and velocity $3\text{ km/hour}$ collides with a rest body of mass $2\text{ m}$ and sticks to it. Now the combined mass starts to move. What will be the combined velocity?
(1996)
Using conservation of linear momentum, $mu_1 = (m+2m)v$, where $u_1 = 3\text{ km/hour}$. Solving gives $3m = 3mv \implies v = 1\text{ km/hour}$
Question 43:
easyTwo 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.
Question 44:
easyTwo 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.
Question 45:
easyWhich 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.
Question 46:
easyTwo particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant, then the speed of centre of mass of the system will be:
(2010 Pre)
Concept: Since external force on the system is zero, the acceleration of the center of mass is zero. Formula: $v_{cm} = \frac{\sum m_i v_i}{\sum m_i}$. Solution: Since the system starts from rest and only internal forces act, velocity of center of mass remains zero.
Question 47:
easyTwo bodies of mass $1text{ kg}$ and $3text{ kg}$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} – 2\hat{j} + \hat{k}$, respectively. The center of mass of this system has a position vector:
(2009)
Concept: Center of mass position vector formula. Formula: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$. Solution: Substituting the given masses and position vectors yields $-2\hat{i} - \hat{j} + \hat{k}$.
Question 48:
easyConsider a system of two particles having masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the mass centre of particles through a distance ‘$d$’ by what distance would the particle of mass $m_2$ move so as to keep the mass centre of particles at the original position:
(2004)
Concept: Shift in center of mass must be zero. Formula: $m_1 \Delta x_1 = m_2 \Delta x_2$. Solution: Substituting $\Delta x_1 = d$ gives $\Delta x_2 = \frac{m_1}{m_2}d$.
Question 49:
easyThe centre of mass of system of particles does not depend on:
(1997)
Concept: Definition and properties of center of mass. Formula: $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
Solution: Center of mass depends only on masses and positions, independent of internal or external forces acting on particles.
Question 50:
easyA uniform circular disc of radius $50\text{ cm}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its center. It is subjected to a torque which produces a constant angular acceleration of $2.0\text{ rad s}^{-2}$. Its net acceleration in $\text{ms}^{-2}$ at the end of $2.0\text{ s}$ is approximately:
(2016-I)
Concept: Combination of tangential and centripetal accelerations. Formula: $a = \sqrt{a_c^2 + a_t^2}$. Solution: $a_t = r\alpha = 1.0$, $a_c = \omega^2 r = 8.0$, giving $a = \sqrt{8^2 + 1^2} \approx 8.0\text{ ms}^{-2}$.