# Impulse and Momentum, Energy Formulations

 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.

 A $19.6\hat{i}$ B $19.6\hat{j}$ C $17.0\hat{i}+9.8\hat{j}$ D $9.8\hat{i}+17.0\hat{j}$
GATE ME 2023   Engineering Mechanics
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
 A 1 kg only B 6 kg only C 1 kg, 6 kg D 1 kg, 9 kg
GATE ME 2019 SET-2   Engineering Mechanics
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

 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
 A 1 B 2 C 4 D $\infty$
GATE ME 2018 SET-1   Engineering Mechanics
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.
 A $\sqrt{\frac{3}{4}}$ B $\sqrt{\frac{3}{2}}$ C 1 D $\sqrt{2}$
GATE ME 2017 SET-1   Engineering Mechanics
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}
 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
 A 4 B 12 C 16 D 24
GATE ME 2017 SET-1   Engineering Mechanics
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.