Tag: momentum

Quizzes (4) Questions (63)

Questions Related to momentum

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

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. $\dfrac{4u}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of angular momentum about the fixed axis A and the coefficient of restitution equation for the impact, the final velocities can be solved. For a rod of mass 3m and length 2l, the moment of inertia about A is (1/3)(3m)(2l)^2 = 4ml^2. Solving the system yields the particle's final speed as u/3.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

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. $9.82rad{s}^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of angular momentum about the center of the disc: L_initial = m*v*r = 0.02 * 5 * 0.1 = 0.01 kg m^2/s. L_final = (I_disc + m*r^2) * omega = (0.5 * 0.1 * 0.1^2 + 0.02 * 0.1^2) * omega = (0.0005 + 0.0002) * omega = 0.0007 * omega. Omega = 0.01 / 0.0007 = 14.2857 rad/s.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

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

By conservation of momentum in the x and y directions: m2*v = m2*(v/2)sin(theta) + m1*v1_x and 0 = m2(v/2)*cos(theta) + m1*v1_y. Since the collision is elastic or specific conditions are implied, the direction of A is determined by the vector sum of momenta. Given the options, the direction is consistent with the conservation laws.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

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 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

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

Multiple choice modelling collisions collisions momentum work, energy and power physics

When a ball collides head-on and elastically with an identical ball on a horizontal frictionless surface,comes to rest while the second one moves with the same velocity as that of the first ball before coillision . 

  1. Can be derived by using momentum conservation alone.

  2. Can be derived by using energy conservation alone

  3. cannot be derived by using any to the two conservation principles.

  4. Can be derived by using both conservation of energy and momentum
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two identical bodies undergo a one-dimensional elastic collision, energy conservation and momentum conservation together are necessary and sufficient to uniquely determine the final velocities, demonstrating that both principles are required.

Multiple choice modelling collisions collisions momentum work, energy and power physics

A ball 'A' of mass 100 gm moving at 2m/s collides with another identical ball 'B' at 3m/s along the same line. There is no loss in energy collision. then, the speed of the balls 'A' and 'B' after the collision are ?

  1. 1m/s and 4m/s

  2. 2m/s and 3 m/s

  3. 3 m/s and 2 m/ s

  4. 2.5 m/s and 2.5 m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an elastic collision between two identical bodies, the velocities are simply exchanged. Ball A initially moving at 2 m/s and Ball B moving at 3 m/s along the same line will swap velocities, meaning Ball A moves at 3 m/s and Ball B moves at 2 m/s after the collision.

Multiple choice modelling collisions collisions momentum work, energy and power physics

Two bodies A and B of masses 5 kg and 10 kg moving in free space in opposite directions with velocity form for second and 0.5 m per second respectively undergo a head on collision the force f of their mutual interaction varies with time T according to the given graph what can you conclude from the given information

  1. Period of Di formation is 0.2 second

  2. Coefficient of restitution is 0.5

  3. Body 0.5 m per second in the original direction

  4. Body be will 1.75 M per second in the reverse direction
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice modelling collisions collisions momentum work, energy and power physics

A solid cylinder of mass 'M' and radius 'R' is rotating along its axis with angular velocity $\omega $ without friction. A particle of mass 'm' moving with velocity v collide against the cylinder and sticks to its rim. After the impact calculate angular velocity of cylinder.

  1. $\cfrac { I+R\omega }{ I+m{ R }^{ 2 } } $
  2. $\cfrac { mvR+IR }{ I+m{ R }^{ 2 } } $
  3. $\cfrac { I\omega +mvR }{ I+m{ R }^{ 2 } }$
  4. $\cfrac { I\omega +mR }{ I+mv{ R }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of angular momentum about the axis of the cylinder: L_initial = I*omega + m*v*R. L_final = (I + m*R^2)*omega_final. Equating them gives omega_final = (I*omega + m*v*R) / (I + m*R^2).