The coefficient of restitution e for a perfectly elastic collision is:
(1988)
By definition, the coefficient of restitution $e$ is equal to $1$ for a perfectly elastic collision where kinetic energy is fully conserved.
Two identical balls $A$ and $B$ having velocities of $0.5\text{ m/s}$ and $-0.3\text{ m/s}$ respectively collide elastically in one dimension. The velocities of $B$ and $A$ after the collision respectively will be:
(2016, 1998, 1994, 1991)
In an elastic collision between two identical bodies, their velocities are mutually exchanged. Given initial velocities are $u_1 = 0.5\text{ m/s}$ and $u_2 = -0.3\text{ m/s}$. Therefore, after collision, the velocities of $B$ and $A$ become $0.5\text{ m/s}$ and $-0.3\text{ m/s}$ respectively.
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: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \varepsilon = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$.
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 $v_2 = \sqrt{v^2 - (v/3)^2} = \frac{2\sqrt{2}}{3}v$.
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 $u_0 = \sqrt{2gh} = 20\text{ m/s}$.
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: $\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$.
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 $\theta = \tan^{-1}(v_{1y}/v_{1x}) = \tan^{-1}(-1/2)$.
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 $\vec{v}_f = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$.
A ball moving with velocity $2\text{ m/s}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$ then their velocities (in $\text{ m/s}$) after collision will be:
(2010 Pre)
Using the collision velocity formulas $v_1 = \frac{(m_1 - em_2)u_1}{m_1+m_2}$ and $v_2 = \frac{(1+e)m_1 u_1}{m_1+m_2}$ with $m_1=m$, $m_2=2m$, $u_1=2$, and $e=0.5$, we get $v_1 = 0$ and $v_2 = 1\text{ m/s}$.
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.