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 $\theta = \tan^{-1}(v_{1y}/v_{1x}) = \tan^{-1}(-1/2)$.
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)
1. $\frac{3}{2}\hat{i} + \frac{1}{4}\hat{j}$
2. $\frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$
3. $\frac{1}{3}v\hat{i} + \frac{2}{3}v\hat{j}$
4. $\frac{2}{3}v\hat{i} + \frac{1}{3}v\hat{j}$
View Answer
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)
1. $0, 2$
2. $0, 1$
3. $1, 1$
4. $1, 0.5$
View Answer
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 $v_2 = 1\text{ m/s}$.