A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is 

(1) $\frac{{\mathrm{Mv}}^2}{\mathrm{T}}$                                 

(2) $\frac{1}{2}\frac{{\mathrm{Mv}}^2}{{\mathrm{T}}^2}$

(3) $\frac{{\mathrm{Mv}}^2}{{\mathrm{T}}^2}$                                   

(4) $\frac{1}{2}\frac{{\mathrm{Mv}}^2}{\mathrm{T}}$

A body of mass 1 kg is thrown upwards with a velocity of $20$ ${\mathrm{ms}}^{-1}.$ It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? $\left(\mathrm{g}=10\right)$ ${\mathrm{ms}}^{-2}$

1. 20 J

2. 30 J

3. 40 J

4. 10 J

The points of maximum and minimum attraction in the curve between potential energy (U) and distance (r) of a diatomic molecules are respectively -

(a) Sand R

(b) T and S

(c) R and S

(d) S and T

K is the force constant of a spring. The work done in increasing its extension from $l_1$ to $l_2$ will be

(1) $K\left(l_2-l_1\right)$                             

(2) $\frac{K}{2}\left(l_2+l_1\right)$

(3) $K\left(l_2^2-l_1^2\right)$                           

(4) $\frac{K}{2}\left(l_2^2-l_1^2\right)$

A body of mass $m_1$ moving with uniform velocity of 40 m/s collides with another mass $m_2$ at rest and then the two together begin to move with uniform velocity of 30 m/s. The ratio of their masses $\frac{m_1}{m_2}$ is

(1) 0.75

(2) 1.33

(3) 3.0

(4) 4.0

A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. The fraction of initial energy transferred to the heavier ball is

(1) $\frac{n}{1+n}$

(2) $\frac{n}{{(1+n)}^2}$

(3) $\frac{2n}{{(1+n)}^2}$

(4) $\frac{4n}{{(1+n)}^2}$

A sphere of mass m moving horizontally with velocity $v_0$ collides against a pendulum bob of mass m. If the two masses stick together after the collision, then the maximum height attained is

(1) $\frac{v_0^2}{2g}$

(2) $\frac{v_0^2}{4g}$

(3) $\frac{v_0^2}{6g}$

(4) $\frac{v_0^2}{8g}$

If a body of mass m moving with velocity u collides head-on elastically  with another identical body at rest. After collision , velocity of the second body will be

(1) zero

(2) u

(3) 2u

(4) data insufficient

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

(a) $\frac{1}{4}\mathrm{v}\hat{\mathbf{i}}+\frac{3}{2}\mathrm{v}\hat{\mathbf{j}}$                                   

(b) $\frac{1}{3}\mathrm{v}\hat{\mathbf{i}}+\frac{2}{3}\mathrm{v}\hat{\mathbf{j}}$

(c) $\frac{2}{3}\mathrm{v}\hat{\mathbf{i}}+\frac{1}{3}\mathrm{v}\hat{\mathbf{j}}$                                   

(d) $\frac{3}{2}\mathrm{v}\hat{\mathbf{i}}+\frac{1}{4}\mathrm{v}\hat{\mathbf{j}}$

Two point masses 1 and 2 move with uniform velocities ${\overrightarrow{\mathrm{v}}}_1 and {\overrightarrow{v}}_2$ respectively. Their initial position vectors are ${\overrightarrow{r}}_1 and {\overrightarrow{r}}_2$ respectively. Which of the following should be satisfied for the collision of the point masses ?

1. $\frac{\overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1}}{\left|\overrightarrow{{\mathrm{r}}_2}+\overrightarrow{{\mathrm{r}}_1}\right|}=\frac{\overrightarrow{v_2}+\overrightarrow{v_1}}{\left|\overrightarrow{v_2}-\overrightarrow{{\mathrm{v}}_1}\right|}$

2. $\frac{\overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1}}{\left|\overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1}\right|}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\left|\overrightarrow{v_2}-\overrightarrow{{\mathrm{v}}_1}\right|}$

3. $\frac{\overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1}}{\left|\overrightarrow{{\mathrm{r}}_2}+\overrightarrow{{\mathrm{r}}_1}\right|}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\left|\overrightarrow{v_2}+\overrightarrow{{\mathrm{v}}_1}\right|}$

4. $\frac{\overrightarrow{{\mathrm{r}}_2}+\overrightarrow{{\mathrm{r}}_1}}{\left|\overrightarrow{{\mathrm{r}}_2}+\overrightarrow{{\mathrm{r}}_1}\right|}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\left|\overrightarrow{v_2}+\overrightarrow{{\mathrm{v}}_1}\right|}$