A metal ball of mass $2\text{ kg}$ moving with speed of $36\text{ km/h}$ has a head on collision with a stationary ball of mass $3\text{ kg}$. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is:
(1997)
1. $100\text{ J}$
2. $140\text{ J}$
3. $40\text{ J}$
4. $60\text{ J}$
View Answer
Initial kinetic energy $K_i = \frac{1}{2}m_1 u_1^2 = 100\text{ J}$ (with $u_1 = 10\text{ m/s}$). Final velocity $v = \frac{m_1 u_1}{m_1+m_2} = 4\text{ m/s}$, and final kinetic energy $K_f = \frac{1}{2}(m_1+m_2)v^2 = 40\text{ J}$. Loss in K.E. = $100 - 40 = 60\text{ J}$.
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)
1. $-0.3\text{ m/s}$ and $0.5\text{ m/s}$
2. $0.3\text{ m/s}$ and $0.5\text{ m/s}$
3. $-0.5\text{ m/s}$ and $0.3\text{ m/s}$
4. $0.5\text{ m/s}$ and $-0.3\text{ m/s}$
View Answer
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.
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)
1. $\frac{\sqrt{3}}{2}v$
2. $\frac{2\sqrt{2}}{3}v$
3. $\frac{3}{4}v$
4. $\frac{3}{\sqrt{2}}v$
View Answer
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)
1. $10\text{ m/s}$
2. $14\text{ m/s}$
3. $20\text{ m/s}$
4. $28\text{ m/s}$
View Answer
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)
1. $\vec{r}_1 - \vec{r}_2 = \vec{v}_1 - \vec{v}_2$
2. $\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$
3. $\vec{r}_1 \cdot \vec{v}_1 = \vec{r}_2 \cdot \vec{v}_2$
4. $\vec{r}_1 \times \vec{v}_1 = \vec{r}_2 \times \vec{v}_2$
View Answer
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)
1. Same as that of B
2. Opposite of that of B
3. $\theta = \tan^{-1}(1/2)$ to the x-axis
4. $\theta = \tan^{-1}(-1/2)$ to the x-axis
View Answer
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 $\theta = \tan^{-1}(v_{1y}/v_{1x}) = \tan^{-1}(-1/2)$.