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Collision of two rigid bodies - class-XI

Description: collision of two rigid bodies
Number of Questions: 22
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Tags: physics momentum work, energy and power systems of particles and rotational motion mechanics
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A uniform rod AB of mass $3m$ and length $2l$ is lying at rest on a smooth horizontal table with a smooth vertical axis through the end $A$ . A particle of mass $2m$ moves with speed $2u$ across the table and strikes the rod at its mid point $C$. If the impact is perfectly elastic , then find the speed of the particle after impact if it strikes the rod normally 

  1. $\dfrac{7u}{3}$

  2. $\dfrac{2u}{3}$

  3. $\dfrac{u}{3}$

  4. $\dfrac{4u}{3}$


Correct Option: C

A body falls from height 20 m.If coefficient of restitution 1/2. The time interval between starting point and second bounce is $\left( {g = 10\;m{s^{ - 2}}} \right)$

  1. 1 s

  2. 2 s

  3. 3 s

  4. 4 s


Correct Option: C

A disc of mass $100g$ and radius $10cm$ has a projection on its circumference. The mass of projection is negligible. A $20g$ bit of putty moving tangential to the disc with a velocity of $5m{s}^{-1}$ strikes the projection and sticks to it. The angular velocity of disc is

  1. $14.29rad{s}^{-1}$

  2. $17.3rad{s}^{-1}$

  3. $12.4rad{s}^{-1}$

  4. $9.82rad{s}^{-1}$


Correct Option: A

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 $\cfrac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

  1. Same as that of $B$

  2. Opposite to that of $B$

  3. $\theta=\tan^{-1}{(1/2)}$ to the x-axis

  4. $\theta=\tan^{-1}{(-1/2)}$ to the x-axis


Correct Option: A

A rod of length on two metal pads of same height from a height $h$. The coefficients of restitution of the metal pads are ${e} _{1}$ and ${e} _{2}$ (${e} _{1}> {e} _{2}$). The angular velocity of the rod after it recoils is

  1. $\cfrac { { e } _{ 1 } }{ { e } _{ 2 } } l\sqrt { 2gh } $

  2. $\cfrac { { e } _{ 1 }-{ e } _{ 2 } }{ l } \sqrt { 2gh } $

  3. $\cfrac { { e } _{ 1 }+1 }{ { e } _{ 2 }+1 } \sqrt { 2gh } $

  4. $\cfrac { { e } _{ 1 }+1 }{ { e } _{ 2 }-1 } \sqrt { 2gh } $


Correct Option: C

In a collision between two solid spheres. velocity of separation along the line of impact (assume no external forces act on the system of two spheres during impact):

  1. Cannot be greater than velocity of approach

  2. Cannot be less than velocity of approach

  3. Cannot be equal to velocity of approach

  4. none of these


Correct Option: A
Explanation:

$\begin{array}{l} e=\dfrac { { volume\, \, of\, \, sep } }{ { volume\, \, of\, \, app } }  \ 0<e<1 \ \Rightarrow Volume\, \, \, of\, \, sep<volume\, \, of\, \, app \end{array}$

$\therefore $ Option $A$ is correct.

Two particles of mass $M _{A} $ and $M _{B} $ and there velocities are $V _{A} $ and $V _{B} $ respectively collides. After collision they inter changes their velocities then ratio of  $\dfrac{M _{A}}{M _{B}}$ is:

  1. (a) $\dfrac{V _{A}}{V _{B}}$

  2. (b) $\dfrac{V _{B}}{V _{A}}$

  3. (c) $\dfrac{V _{A}+V _{B}}{V _{B}-V _{A}}$

  4. (d) 1


Correct Option: A
Explanation:
The correct option is B.

Given


Two particles having mass $M _a \& M _b$ and the velocities are $V _a\$V_b$

so when the collides then they interchange their velocities.

Thus the ratio of their velocities after collission is :

$\dfrac{V_a}{V_b}$

A plastic ball falls from a height of $4.9$ metre and rebounds several times from the floor. What is the coefficient of restitution during the impact with the floor if $1.3$ seconds pass from the first impact to the second one?

  1. $0.9$

  2. $0.1$

  3. $0.7$

  4. $0.8$


Correct Option: B
Explanation:
As, $ t = \sqrt{\dfrac{2h}{g}} $
Velocity of ball just before collision $ = v = \sqrt{2gh} $
After first collision,
Velocity $ = v _{1} = ev = e\sqrt{2gh} $
So, $ t _{1} = \dfrac{v _{1}}{g} = e \sqrt{\dfrac{2h}{g}} $
So, time for second collision will be 
$ T = t+2t _{1} $
$ = (1+2e)\sqrt{\dfrac{2h}{g}} $
Putting, $ T = 1.3\,sec.,h = 4.9\,m $
$ 1.3 = (1+2e)\sqrt{\dfrac{2\times 4.9}{9.8}} $
$ 1.3 = 1+2e $
$ 0.3 = 2e $
$ \boxed{e = 0.1} $ 

As shown in figure a particle is released from highest point of curved smooth path. Find the distance of point of strike from $A$.

  1. $10\ m$

  2. $5\ m$

  3. $20\ m$

  4. $none\ of\ these$


Correct Option: C

A body 'x' with a momentum 'p' collides with with another identical stationary body 'y' dimensionally. During the collision 'y' gives an impulse 'J' to the body 'x'. Then the coefficient of restitution is 

  1. $\dfrac p{p-2J}$

  2. $\dfrac p{p-J}$

  3. $\dfrac p{p+2J}$

  4. $\dfrac p{p+J}$


Correct Option: A

A uniform rod AB of length $L$ and mass $M$ is lying on a smooth table. A small particle of mass $m$ strike the rod with a velocity $v _0$ at point C a distance x from the centre O. The particle comes to rest after collision. The value of $x$, so that point A of the rod remains stationary just after the collision,  is:

  1. $L/3$

  2. $L/6$

  3. $L/4$

  4. $L/12$


Correct Option: B

a body of mass m falls from height h on ground. If e be the coefficient of restitution of collision betwwen the body and ground then the distance travelled by body before it comes to rest is

  1. $h\left{ {\dfrac{{1\, + {e^2}}}{{1 - {e^2}}}} \right}$

  2. $h\left{ {\dfrac{{1\, - {e^2}}}{{1 + {e^2}}}} \right}$

  3. ${\dfrac{{2eh}}{{1 + {e^2}}}}$

  4. ${\dfrac{{2eh}}{{1 - {e^2}}}}$


Correct Option: C

A solid spherical ball of radius R collides with a rough horizontal surface as shown in figure. At the time of collision its velocity is $v _{0}$ at an angle $\theta$ to the horizontal and angular velocity $\omega _{0}$ as shown. After collision, angular velocity of ball may

  1. decrease

  2. increase

  3. remains constant

  4. none of these


Correct Option: A

A particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$. When it moves from one point to a diametrically opposite point, its:

  1. momentum does not change

  2. momentum changes by $2Mv$

  3. $KE$ changes by $Mv^{2}$

  4. none of the above


Correct Option: B

If a ball is dropped from rest, it bounces from the floor. The coefficient of restitution is $0.5$ and the speed just before the first bounce is $5\ m/sec$. The total time taken by the ball to come to rest is:

  1. $2\ sec$

  2. $1\ sec$

  3. $0.5\ sec$

  4. $0.25\ sec$


Correct Option: C

A solid sphere rolls without slipping on a rough horizontal floor, moving with a speed $v$. It makes an elastic collision with a smooth vertical wall. After impact,

  1. it will move with a speed $v$ initially.

  2. its motion will be rolling without slipping.

  3. its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant.

  4. its motion will be rolling without slipping only after some time.


Correct Option: C,D
Explanation:

The velocity would abruptly change and would cause the solid sphere to slide at first and after some time it would attain a constant angular velocity where it would roll without slipping.

An athelete diving off a high spring board can perform a variety of physical moments in the air before entering the water below. Which one of the following parameters will remain constant during the fall? The athelete's:

  1. linear velocity

  2. linear momentum

  3. moment of inertia

  4. angular velocity


Correct Option: D

In an elastic collision, kinetic energy of the relative motion is converted into the ____ energies of two momentarily compressed bodies, and then is converted back into the _____ energy. Fill in the blanks. 

  1. kinetic,kinetic

  2. elastic,kinetic

  3. elastic,elastic

  4. kinetic,elastic


Correct Option: B
Explanation:

During elastic collision, at first both the bodies get deformed, so total kinetic energy is converted into theie elastic energies and then they regain their original shape (when moving apart with different velocities) suggesting that the elastic energy is converted back into the kinetic energy.

State whether the given statement is True or False :

In an elastic collision, the net kinetic energy of the two colliding bodies is conserved.

  1. True

  2. False


Correct Option: A
Explanation:

In an elastic collision, the net kinetic energy as well as linear momentum of the two colliding bodies is conserved. The given statement is true.

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

  1. $\dfrac{1-e}{1+e}$

  2. $\dfrac{e-1}{e+1}$

  3. $\dfrac{1+e}{1-e}$

  4. $\dfrac{e+1}{e-1}$


Correct Option: A
Explanation:

A. $\dfrac{1-e}{1+e}$


Given,


$m _1=m _2=m$ (say)

$u _1=u $, $u _2=0$

let, $v _1=$ velocity of ball 1 after collision

      $v _2=$ velocity of ball 2 after collision

The coefficient of restitution,

$e=\dfrac{v _2-v _1}{u _1-u _2}$

$eu=v _2-v _1$. . . . . . .(1)

By the conservation of Linear momentum,

$m _1u _1+m _2u _2=m _1v _1+m _2v _2$ 

$u=v _1+v _2$. . . . . . . . .(1)

By solving equation (1) and (2), we get

$v _1=\dfrac{(1-e)u}{2}$

$v _2=\dfrac{(1+e)u}{2}$

The ratio of velocity of two ball,

$\dfrac{v _1}{v _2}=\dfrac{1-e}{1+e}$

A sphere $A$ moving with speed $u$ and rotating with an angular velocity $\omega$ makes a head-on elastic collision with an identical stationary sphere $B$. There is no friction between the surfaces of $A$ and $B$. Choose the correct alternative(s). Discard gravity.

  1. $A$ will stop moving but continue to rotate with an angular velocity $\omega$

  2. $A$ will come to rest and stop rotating

  3. $B$ will move with speed $u$ without rotating

  4. $B$ will move with speed $u$ and rotate with an angular velocity $\omega$.


Correct Option: A,C
Explanation:
Let $m$ be the mass of sphere and $v _1$ and $v _2$ be the velocities of the spheres A and B respectively after the collision.
Applying conservation of linear momentum:        $P _i = P _f$
$m  u + 0 = mv _1  + mv _2           \implies v _1 + v _2 = u$      ........(1)
Also $\dfrac{v _2- v _1}{ u-0} = -1         \implies v _1 - v _2 = -u$  
On solving we get,       $v _1 = 0 $   and $v _2 = u$
Thus A will stop and B will move with $u$
Also as there is no friction between the A and B, thus there will be no torque. Hence angular velocities of the respective spheres must remains the same as it was initially.
So, A will continue to rotate with $w$ whereas B will not rotate.

A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

  1. 8.4m/s

  2. 4.2m/s

  3. 2.1m/s

  4. 16.8m/s


Correct Option: A
Explanation:

At first we can find the final velocity of the $100\ g$ ball.

$ \dfrac { { m } _{ 2 }{ { v } _{ 2 } }^{ 2 } }{ 2 } =mgh={ m } _{ 2 }gl(1-cos\alpha )$

$ \implies { v } _{ 2 }=\sqrt { 2gl\left( 1-cos\alpha  \right)  } =\sqrt { 2*9.81*1.1(1-cos{ 50 }^{ o }) } =2.78m/s $

Now applying conservation of the momentum principle, we get:

$ { m } _{ 1 }{ V } _{ 1 }+{ m } _{ 2 }{ V } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$

$ { v } _{ 1 }=\dfrac { { m } _{ 1 }{ V } _{ 1 }-{ m } _{ 2 }{ v } _{ 2 } }{ { m } _{ 1 } } ={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } $

Principle of kinetic energy conservation , we get 

$ \dfrac { 1 }{ 2 } { m } _{ 1 }{ { { V } _{ 1 } }^{ 2 } }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { V } _{ 2 } }^{ 2 }=\dfrac { 1 }{ 2 } { m } _{ 1 }{ { v } _{ 1 } }^{ 2 }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { v } _{ 2 } }^{ 2 }$


$ { v } _{ 1 }={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } \ { V } _{ 2 }=0$

$\implies { V } _{ 1 }=\dfrac { { { { { m } _{ 1 } }^{ 2 }v } _{ 1 } }^{ 2 }+{ m } _{ 2 }{ m } _{ 1 }{ { v } _{ 2 } }^{ 2 } }{ 2{ m } _{ 1 }{ v } _{ 2 }{ m } _{ 2 } } =\dfrac { v _{ 2 }\left( { m } _{ 2 }+{ m } _{ 1 } \right)  }{ 2{ m } _{ 1 } } $

$ \implies  \dfrac { 2.78\left( 0.1+0.024 \right)  }{ 2*0.024 } =8.4m/s$

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