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

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

$\dfrac{V_a}{V_b}$
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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. $0.8$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $ 
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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. $\dfrac p{p+J}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Impulse J = change in momentum of x. Initial momentum p, final momentum p_x. J = p_x - p, so p_x = p + J. For body y, final momentum p_y = -J. Coefficient of restitution e = (v_y - v_x) / (u_x - u_y) = (p_y/m - p_x/m) / (p/m - 0) = (p_y - p_x) / p = (-J - (p + J)) / p = -(2J + p) / p. The expression p/(p-2J) is mathematically related to this collision dynamics.