Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions | Engineering Mechanics | Applied Mechanics and Design | Mechanical | GATE Syllabus, Paper Solution, Question Answer

Weightage of Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Total 11 Questions have been asked from Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions topic of Engineering Mechanics subject in previous GATE papers. Average marks 1.55.

Question No. 38

GATE - 2018
02

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 10 $ m/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 $

Show Answer

(C) 4

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

$ \begin{array}{l}\mathrm V^2=\mathrm u^2+2\mathrm{gh}(2^\mathrm{nd}\;\mathrm{law}\;\mathrm{of}\;\mathrm{motion})\\\mathrm V=0,\mathrm u=4\;\mathrm m/\sec\\\mathrm g=-10\;\mathrm m/\sec^2(\mathrm{Travelling}\;\mathrm{against}\;\mathrm{gravity})\\0=\mathrm y^2-2\times10\times\mathrm H\\\mathrm H=0.8\;\mathrm m\\\mathrm V=\mathrm u+\mathrm{gt}\\4={\mathrm{gt}}_1\\{\mathrm t}_1=0.4\end{array} $

So, time taken before 1^st collsion $ =2\times0.4=85 $

after 1^st collsion, the velocity becomes

$ 0.8\times{\mathrm V}_1=0.8\times4=3.2\;\mathrm m/\sec $

Time travelled in the air between 1^st & 2^nd codition

$ =2\times{\mathrm t}_1=2\times0.32 $

After 2^nd collsion the velocity becomes

$ 0.8^2\times4=2.56\;\mathrm m/\sec $

Time travelled in the air between 2^nd & 3^rd collision

$ =2\times0.256 $

$ \begin{array}{l}\mathrm{Total}\;\mathrm{time}=2\times0.4+2+0.32+2\times0.32+2\times0.256+.........\infty\;\\2\lbrack0.4+0.32+0.256+........\infty\rbrack\\2\left[\frac{0.4}{1-{\displaystyle\frac45}}\right]\\\mathrm T\Rightarrow4\;\sec\\\mathrm a=4\\\mathrm r=\frac{0.32}{0.4}=\frac45\\{\mathrm s}_{\circ\circ}=\frac4{1-\mathrm r}\\\;\;\downarrow\end{array}$

Sum of infinite terms (G,P)

Question No. 40

GATE - 2017
02

A point mass of 100 kg is dropped onto a massless elastic bar (cross-sectional area = 100 mm², length = 1 m, Young's moduls = 100 GPa) from a height H of 10 mm as shown (Figure is not to scale). If g = 10 m/s², the maximum compression of the elastic bar is ____________mm.

Show Answer

1.50 to 1.52

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

For maximum compression

Potential energy of mass 'm' must be conserved as elastic energy of rod when 'm' comes to rest. stiffness of bar

$ =\mathrm k=\frac{\mathrm{AE}}{\mathrm I}=\frac{100\times10^{-6}\times100\times10^9}{1000}=10^4\;\mathrm n/\mathrm{mm} $

Let x is the compression. (in mm)

Thus mg $ \left(\mathrm H+\mathrm x\right)=\frac12\mathrm{kx}^2 $

$ 100\times10\times\left(10+\mathrm x\right)=\frac12\times10^4\;\mathrm x^2 $

$ 10+\mathrm x=5\mathrm x^2 $

$ 5\mathrm x^2-\mathrm x-10=0 $

$ \mathrm x=\frac{-1\pm\sqrt{1+200}}{2\left(5\right)}=\frac{1\pm\sqrt{201}}{10} $

$ \mathrm x=\frac{1+\sqrt{201}}{10}=1.517\;\mathrm{mm} $

Question No. 116

GATE - 2016
01

A point mass having mass $ M $ is moving with a velocity $ V $ at an angle $ \theta $ to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is

(A) $-2MV\;\cos\;\theta\;\widehat j$
(B) $2MV\;\sin\;\theta\;\widehat j$
(C) $2MV\;\cos\;\theta\;\widehat j$
(D) $-2MV\;\sin\;\theta\;\widehat j$

Show Answer

(D) $-2MV\;\sin\;\theta\;\widehat j$

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

Adter elastic collision , ball will rebound with same vertical speed as that before impact.

Total energy in momentum = Final momentum - Initial momentum $ =-\mathrm m\;\mathrm V\;\mathrm{sinθ}\;\mathrm j-\mathrm m\;\mathrm V\;\mathrm{sinθ}\;\mathrm j\;=0-2\mathrm{mvsinθ}\;\widehat{\mathrm j} $

Question No. 140

GATE - 2016
02

A mass of 2000 kg is currently being lowered at a velocity of 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is $150\mathrm{kg}-\mathrm m^2$. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is __________

Show Answer

14.1 : 14.3

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

$ \begin{array}{l}\therefore\mathrm V=2\;\mathrm m/\sec\\\therefore\mathrm V=\mathrm{rω}\\\mathrm\omega=\frac{\mathrm v}{\mathrm r}=\frac21=2\;\frac{\mathrm{rad}}\sec\\{\left({\mathrm E}_\mathrm{abs}\right)}_\mathrm{brake}=\frac12\vert{\mathrm\omega}_1^2+\frac12{\mathrm{mv}}_1^2+\mathrm{mg}\left({\mathrm H}_1-{\mathrm H}_2\right)\\{\left({\mathrm E}_\mathrm{abs}\right)}_\mathrm{brake}=\begin{bmatrix}\frac12\times150\times2^2+\frac12\times2000\\\times2^2+2000\times9.81\times0.5\;\end{bmatrix}\\{\left({\mathrm E}_\mathrm{abs}\right)}_\mathrm{brake}=14110\;\mathrm J=14.11\;\mathrm{KJ}\\\\\end{array} $

Question No. 141

GATE - 2016
02

A system of particles in motion has mass center $G$ as shown in the figure. The particle $i$ has mass $m_i$ and its position with respect to a fixed point $O$ is given by the position vector $ {\boldsymbol r}_\mathbf i $. The position of the particle with respect to $G$ is given by the vector $ {\boldsymbol\rho}_\mathbf i $. The time rate of change of the angular momentum of the system of particles about $G$ is

(The quantity $ {\overset{\boldsymbol.\boldsymbol.}{\mathbf\rho}}_\mathbf i\boldsymbol\infty $ indicates second derivative of $ {\boldsymbol p}_\mathbf i $ with respect to time and likewise for $ \boldsymbol r\boldsymbol i$ ).

(A) $ {\textstyle\sum_i}r_i\times m_i\overset{..}{{\boldsymbol p}_i\;} $
(B) $ {\textstyle\sum_i}\rho_i\times m_i{\overset{\boldsymbol.\boldsymbol.}{\mathbf r}}_i $
(C) $ {\textstyle\sum_i}r_i\times m_i{\overset{\boldsymbol.\boldsymbol.}{\mathbf r}}_i $
(D) $ {\textstyle\sum_i}\rho_i\times m_i{\overset{..}{\boldsymbol p}}_i $

Show Answer

(B) $ {\textstyle\sum_i}\rho_i\times m_i{\overset{\boldsymbol.\boldsymbol.}{\mathbf r}}_i $

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

Angular momentum of system of particles about G is

$ \mathrm H={\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i\dot{{\mathrm r}_\mathrm i} $

Time rate of change of angular momentum about G is

$ \begin{array}{l}\frac{\mathrm{dH}}{\mathrm{dt}}=\frac{\mathrm d}{\mathrm{dt}}\left({\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i\dot{{\mathrm r}_\mathrm i}\right)\\\mathrm H=\mathrm\Sigma\left(\dot{{\mathrm P}_\mathrm i}\times{\mathrm m}_\mathrm i\dot{{\mathrm r}_\mathrm i}\right)+\left({\mathrm P}_\mathrm i\times{\mathrm m}_\mathrm i\dot{{\mathrm r}_\mathrm i}\right).......(1)\end{array} $

For geometry, by vector addition:

$ {\mathrm r}_\mathrm i={\mathrm P}_\mathrm i+{\mathrm r}_0.......(2) $

On differentiation,

$ \begin{array}{l}{\dot{\mathrm r}}_\mathrm i={\dot{\mathrm P}}_\mathrm i+{\mathrm r}_0.......(3)\\\ddot{{\mathrm r}_\mathrm i}={\ddot{\mathrm P}}_\mathrm i+{\ddot{\mathrm r}}_0.......(4)\end{array} $

Using 3 in 1,

$ \begin{array}{l}\dot{\mathrm H}=\mathrm\Sigma\left({\dot{\mathrm r}}_\mathrm i-{\dot{\mathrm r}}_0\right)\times{\mathrm m}_\mathrm i{\dot{\mathrm r}}_\mathrm i+{\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i{\ddot{\mathrm r}}_\mathrm i\\\dot{\mathrm H}=\mathrm\Sigma{\dot{\mathrm r}}_\mathrm i\times{\mathrm m}_\mathrm i{\dot{\mathrm r}}_\mathrm i-\mathrm\Sigma{\dot{\mathrm r}}_0\times\left(\mathrm m{\dot{\mathrm r}}_\mathrm i\right)+{\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i{\ddot{\mathrm r}}_\mathrm i\\\dot{\mathrm H}=\mathrm\Sigma{\dot{\mathrm r}}_\mathrm i\times{\mathrm m}_\mathrm i{\dot{\mathrm r}}_\mathrm i=0\;\;\;\;\;\;(\mathrm{Cross}\;\mathrm{product}\;\mathrm{of}\;\mathrm{similar}\;\mathrm{qualities})\\\mathrm\Sigma{\dot{\mathrm r}}_0\times\left(\mathrm m{\dot{\mathrm r}}_\mathrm i\right)={\dot{\mathrm r}}_0\times\left({\mathrm{Σm}}_\mathrm i\dot{{\mathrm r}_\mathrm i}\right)={\dot{\mathrm r}}_0\times\left(\mathrm m{\dot{\mathrm r}}_0\right)=0\end{array} $

[Using definition of centre of mass]

$ \begin{array}{l}\dot{\mathrm H}={\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i{\ddot{\mathrm r}}_\mathrm i.....................(\mathrm b)\\\end{array} $

One can also use equation 4,

$ \begin{array}{l}\dot{\mathrm H}={\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i\left(\ddot{{\mathrm P}_\mathrm i}+{\ddot{\mathrm r}}_0\right)\\={\mathrm{ΣP}}_\mathrm i\times\left({\mathrm m}_\mathrm i\ddot{{\mathrm P}_\mathrm i}\right)+{\mathrm{ΣP}}_\mathrm i\times{\mathrm m}_\mathrm i{\ddot{\mathrm r}}_0\\={\mathrm{ΣP}}_\mathrm i\times\left({\mathrm m}_\mathrm i\ddot{{\mathrm P}_\mathrm i}\right)+\mathrm\Sigma\left({\mathrm m}_\mathrm i{\mathrm P}_\mathrm i\right)\times{\ddot{\mathrm r}}_0\\={\mathrm{ΣP}}_\mathrm i\times\left({\mathrm m}_\mathrm i\ddot{{\mathrm P}_\mathrm i}\right)+0\\({\mathrm{Σm}}_\mathrm i{\mathrm P}_\mathrm i=0\Rightarrow\mathrm{definition}\;\mathrm{center}\;\mathrm{of}\;\mathrm{mass})\\\dot{\mathrm H}={\mathrm{ΣP}}_\mathrm i\times\left({\mathrm m}_\mathrm i\ddot{{\mathrm P}_\mathrm i}\right).........................(\mathrm d)\end{array} $

Question No. 50

GATE - 2015
02

A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20% . The height (in m) to which the ball will rise is _____.

Show Answer

0.64

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

. during impact $m_1v_1$ = $m_2v_2$

but here $m_1$ = $m_2$ $\therefore$ $v_1$ = $v_2$

Now v = $\begin{array}{l}\sqrt{2gh}\\\end{array}$ also $v_1$ = 0.8 $v_1$

$\begin{array}{l}h=\frac{v^2}{2g}\\\end{array}$

$h_1$ = $\begin{array}{l}v_1^2/2g\\\end{array}$ $h_2$ = $\begin{array}{l}v_2^2/2g\\\end{array}$

= $(0.8 v_1)^2/2g$

= 0.64 $(v_1^2/2g)$

= 0.64 $h_1$

= 0.64 x 1

= 0.64 m .

Question No. 116

GATE - 2015
01

A small ball of mass 1 kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be____.

Show Answer

7.8 to 8.2

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Explanation :

For elastic collision

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\;\;\_\_(1)moment\;conservation$

$(1\times12)+(2\times0)=(1\cdot v_1)+(2\cdot v_2)$

$12+0=v_1+2v_2\;\;\_\_(2)$

$1/2\;m_1u_1^2+1/2\;m_2u_2^2=1/2\;m_1v_1^2+1/2\;m_2v_2^2\;\_\_(3)$

$(1/2\cdot1\cdot12^2)+(1/2\cdot2\cdot0^2)=(1/2\cdot1\cdot v_1^2)+(1/2\cdot2\cdot v_2^2)$

$144+0=v_1^2+2v_2^2\;\;\_\_(4)$

$From\;(2)\;\;v_1=12-2v_2$

$144=(12-2v_2)^2+2v_2^2$

$144=144-48v_2+4v_2+2v_2^2$

$6v_2^2-48v_2=0$

$6v_2(v_2-8)=0$

$\therefore\;v_2=8\;m/sec$

Question No. 219

GATE - 2014
01

A mass m₁ of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with a stationary mass m₂ of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is

(A) 0.6
(B) 0.1
(C) 0.01
(D) 0

Show Answer

(D) 0

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Question No. 22

GATE - 2011
01

The coefficient of restitution of a perfectly plastic impact is

(A) 0
(B) 1
(C) 2
(D) ∞

Show Answer

(A) 0

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Question No. 32

GATE - 2011
02

A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force F_x (in N) exerted by the rubber band as a function of distance x (in m) is given by F_x = 300x². If the stone is displaced by 0.1m from the un-stretched position (x = 0) of the rubber band, the energy stored in the rubber band is

(A) 0.01 J
(B) 0.1 J
(C) 1 J
(D) 10 J

Show Answer

(B) 0.1 J

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Question No. 12

GATE - 2007
01

During inelastic collision of two particles, which one of the following is conserved?

(A) total linear momentum only
(B) total kinetic energy only
(C) both linear momentum and kinetic energy
(D) neither linear momentum nor kinetic energy

Show Answer

(A) total linear momentum only

Subject : Engineering Mechanics

Topic : Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Questions & Answers of Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

Weightage of Impulse and Momentum (Linear and Angular) and Energy Formulations, Collisions

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(C) 4

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1.50 to 1.52

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(D) $-2MV\;\sin\;\theta\;\widehat j$

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14.1 : 14.3

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(B) $ {\textstyle\sum_i}\rho_i\times m_i{\overset{\boldsymbol.\boldsymbol.}{\mathbf r}}_i $

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0.64

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7.8 to 8.2

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(D) 0

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(A) 0

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(B) 0.1 J

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(A) total linear momentum only

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