Question 1 |

A spherical ball weighing 2 kg is dropped from a height of 4.9 m onto an
immovable rigid block as shown in the figure. If the collision is perfectly elastic,
what is the momentum vector of the ball (in kg m/s) just after impact?

Take the acceleration due to gravity to be g=9.8 m/s^2. Options have been rounded off to one decimal place.

Take the acceleration due to gravity to be g=9.8 m/s^2. Options have been rounded off to one decimal place.

19.6\hat{i} | |

19.6\hat{j} | |

17.0\hat{i}+9.8\hat{j} | |

9.8\hat{i}+17.0\hat{j} |

Question 1 Explanation:

Let '\theta' be the angle about line of impact through which ball will move often the impact.

Let 'u' be the vertical downward velocity of the ball before striking and 'v' be the velocity of ball after the impact which make an angle '\theta' with the line of impact. As ball fall freely under the gravity from height \mathrm{h}=4.9 \mathrm{m}, hence downward velocity '\mathrm{u}' at the instance of striking the rigid body

u=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 4.9}

or, u=9.8 \mathrm{~m} / \mathrm{sec}

for perfectly elastic collision, e=1

As e=\frac{\text { relative velocity of seperation }}{\text { relative velocity of approach }}=1\;\;...(i)

As block is rigid, so block velocity =0.

So, along the line of impact,

relative velocity of approach =u \cos 30^{\circ}-0 =u \cos 30^{\circ}

relative velocity of separation along the line of impact =v \cos \theta-0=v \cos \theta

so, by equation (i)

u \cos 30^{\circ}=v \cos \theta \;\;\;...(ii)

In the direction normal to the impact, the component velocity is not affected so,

u \sin 30^{\circ}=v \sin \theta \;\;\;...(iii)

So, by equation (ii) and (iii)

V=\sqrt{u^{2} \cos ^{2} 30^{\circ}+u^{2} \sin ^{2} 30^{\circ}}

or, V=9.8 \mathrm{~m} / \mathrm{sec}

by equation (ii) and (iii)

\tan \theta=\tan 30^{\circ}, \theta=30^{\circ}

So, inclination to the plane for of the section

=90^{\circ}-30^{\circ}=60^{\circ}

So, momentum equation is given by during seperation:

\vec{P}_{s}=m v \cos 30^{\circ} \hat{i}+m v \cos 30^{\circ} \hat{j} =17 \hat{i}+9.8 \hat{j}

Question 2 |

A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value(s) of m is/are

1 kg only | |

6 kg only | |

1 kg, 6 kg | |

1 kg, 9 kg |

Question 2 Explanation:

Let V_{1} is the speed of 3 kg mass after collision

\mathrm{V}_{2} is the speed of m kg mass after collision

\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}

By linear momentum conservation

\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}

By using (1), (2) \& (3) we get

m = 1 kg (or) 9 kg

\mathrm{V}_{2} is the speed of m kg mass after collision

\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}

By linear momentum conservation

\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}

By using (1), (2) \& (3) we get

m = 1 kg (or) 9 kg

Question 3 |

A point mass is shot vertically up from ground level with a velocity of 4 m/s at time, t_0. It loses 20% of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is 10m/s^{2} and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of

1 | |

2 | |

4 | |

\infty |

Question 3 Explanation:

\begin{aligned} (1) \rightarrow \qquad t&=? \\ v&=u+a t \\ 0&=4-10 t \\ t&=\frac{4}{10}=0.4 s \\ (2) \rightarrow \qquad t^{\prime}&=? \\ u^{\prime}&=0.8 \times u \\ &=0.8 \times 4=3.2 \mathrm{m} / \mathrm{s} \\ \mathrm{v}^{\prime}&=U^{\prime}+\mathrm{al}^{\prime} \\ 0&=3.2-10 t^{\prime} \\ t^{\prime}&=\frac{3.2}{10}=0.32 \mathrm{s} \\ (3) \rightarrow \qquad t^{\prime \prime}&=? \\ U^{\prime \prime} &=0.8 u^{\prime} \\ &=0.8 \times 3.2=2.56 \mathrm{m} / \mathrm{s} \\ V^{\prime \prime} &=U^{\prime \prime}+a t^{\prime} \\ 0 &=2.56-10 t^{\prime \prime} \\ t^{\prime \prime} &=0.256 s \end{aligned}

So, t, t^{\prime}, t^{\prime \prime} are forming a GP series

\begin{array}{l} \text{So, total time }=2\left(t+t^{\prime}+t^{\prime \prime}+\ldots . .0\right)\\ =2[0.4+0.32+0.256+\ldots .0] \\ =2 \times \frac{0.4}{1-0.8}=2 \times 2=4 s \end{array}

Question 4 |

Two disks A and B with identical mass (m) and radius (R) are initially at rest. They roll down from the top of identical inclined planes without slipping. Disk A has all of its mass concentrated at the rim, while Disk B has its mass uniformly distributed. At the bottom of the plane, the ratio of velocity of the center of disk A to the velocity of the center of disk B is.

\sqrt{\frac{3}{4}} | |

\sqrt{\frac{3}{2}} | |

1 | |

\sqrt{2} |

Question 4 Explanation:

m_{A}=m_{B}=m

Outer circumferential radius of both disc A

and B is also same i.e. R

\begin{aligned} I_{A}&=m R^{2} \\ I_{B}&=\frac{m R^{2}}{2} \end{aligned}

\frac{V_{A}}{V_{B}} at bottom of plane =?

m g h=(K \cdot E) of disc A=K \cdot E of disc B

\begin{aligned} \frac{1}{2} m V_{A}^{2}+\frac{1}{2} I_{A} \omega_{A}^{2} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} I_{B} \omega_{B}^{2} \\ \frac{1}{2} m V_{A}^{2}+\frac{1}{2} m R^{2} \frac{V_{A}^{2}}{R^{2}} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} \frac{m R^{2}}{2} \times \frac{V_{B}^{2}}{R^{2}} \\ \frac{V_{A}^{2}}{2}+\frac{V_{A}^{2}}{2} &=\frac{V_{B}^{2}}{2}+\frac{V_{B}^{2}}{4} \\ V_{A}^{2} &=\frac{3}{4} V_{B}^{2} \\ \frac{V_{A}}{V_{B}} &=\sqrt{\frac{3}{4}} \end{aligned}

Outer circumferential radius of both disc A

and B is also same i.e. R

\begin{aligned} I_{A}&=m R^{2} \\ I_{B}&=\frac{m R^{2}}{2} \end{aligned}

\frac{V_{A}}{V_{B}} at bottom of plane =?

m g h=(K \cdot E) of disc A=K \cdot E of disc B

\begin{aligned} \frac{1}{2} m V_{A}^{2}+\frac{1}{2} I_{A} \omega_{A}^{2} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} I_{B} \omega_{B}^{2} \\ \frac{1}{2} m V_{A}^{2}+\frac{1}{2} m R^{2} \frac{V_{A}^{2}}{R^{2}} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} \frac{m R^{2}}{2} \times \frac{V_{B}^{2}}{R^{2}} \\ \frac{V_{A}^{2}}{2}+\frac{V_{A}^{2}}{2} &=\frac{V_{B}^{2}}{2}+\frac{V_{B}^{2}}{4} \\ V_{A}^{2} &=\frac{3}{4} V_{B}^{2} \\ \frac{V_{A}}{V_{B}} &=\sqrt{\frac{3}{4}} \end{aligned}

Question 5 |

A particle of unit mass is moving on a plane. Its trajectory, in polar coordinates, is given by r(t)=t^{2},\theta (t)=t, where t is time. The kinetic energy of the particle at time t=2 is

4 | |

12 | |

16 | |

24 |

Question 5 Explanation:

\begin{aligned} r &=t^{2}: \theta=t \\ \mathrm{K.E} &=\frac{1}{2} m v^{2}=? \text { at } t=2 \mathrm{sec} \\ \Rightarrow m &=1 \mathrm{kg} \\ \vec{V} &=r \omega(\hat{t})+\frac{d r}{d t} \dot{r}=t^{2} \times 1(\hat{t})+2 t \hat{r} \\ \therefore \frac{d \theta}{d t}&=\omega=1
\\ \vec{V} &=t^{2}(\hat{t})+2 t(\hat{r}) \\ \text{at }\qquad t &=2 s \\ \vec{V} &=4(\hat{t})+4(\hat{r}) \\ |\vec{V}| &=\sqrt{16+16}=\sqrt{32} \\ K.E. &=\frac{1}{2}mv^{2}=\frac{1}{2}\times1\times32=16 \end{aligned}

There are 5 questions to complete.