Tag: work, energy and power

Quizzes (16) Questions (235)

Questions Related to work, energy and power

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

Two equal masses ${m} _{1}$ and ${m} _{2}$ moving along the same straight line with velocities $+3  {m}/{s}$ and $-5  {m}/{s}$ respectively collide elastically. Their velocities after the collision will be respectively :

  1. $+4 {m}/{s}$ for both
  2. $-3 {m}/{s}$ and $+5 {m}/{s}$
  3. $-4 {m}/{s}$ and $+4 {m}/{s}$
  4. $-5 {m}/{s}$ and $+3 {m}/{s}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As ${m} _{1} = {m} _{2}$, therefore after elastic collision velocities of masses gets interchanged.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

A machine gun of mass 5 kg fires 30 bullets, each of mass 50 g, per minute at a speed of $400\,m\,{s^{ - 1}}$. What force must be exerted to keep the machine gun in position?

  1. $30N$
  2. $10N$
  3. $27N$
  4. $33N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mass of the gun $=5kg$

Mass of bullets $=50g$
speed at which the bullet is fired$=400m/s$
Therefore$,$ force required to keep the machine gun in position $=\frac{{ch\arg e\,in\,momentum\,of\,bullets}}{{time\,taken}}$
$=\frac{{\left( {50 \times {{10}^{ - 3}} \times 30} \right) \times 400}}{{60}} = 10\,N$
Hence,
option $(B)$ is correct answer.

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.

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.