Questions Related to physics

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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. $L/12$

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For point A to be stationary, the velocity of the center of mass v_cm and the angular velocity omega must satisfy v_A = v_cm - omega * (L/2) = 0. Using conservation of linear and angular momentum: mv0 = M*v_cm and mv0*x = I*omega = (ML^2/12)*omega. Substituting v_cm = mv0/M and omega = 12mv0x/ML^2 into the condition gives x = L/6.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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

Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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

Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial momentum vector is Mv at an angle. Final momentum vector at the diametrically opposite point is -Mv. The change in momentum is final - initial = -Mv - Mv = -2Mv. The magnitude of the change is 2Mv.

Multiple choice physics systems of particles and rotational motion collision of two rigid bodies energy and collisions understanding collisions

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.

Reveal answer Fill a bubble to check yourself
C,D Correct answer
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.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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

Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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

Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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