Tag: work, energy and power

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Questions Related to work, energy and power

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A lighter body moving with a velocity $v$ collides with a heavier body at rest. Then :

  1. the lighter body rebounces with twice the velocity of bigger body

  2. the lighter body retraces its path with the same velocity in magnitude

  3. the heavier body does not move practically

  4. both (2) and (3)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a collision system where $m _1$ moves with $u _1$ initially and $m _2$ is at rest with $m _2 >>> m _1$.
Let the final velocity of $m _1$ be $v _1$ and $m _2$ be $v _2$.
Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,
$ m _1u _1 = m _2v _2 + m _1v _1 $
$ v _2 - v _1 = u _1 $


We get $ v _1 = \dfrac {m _1 - m _2 }{m _1 + m _2} u _1 $

$ v _2 = \dfrac {2m _1}{m _1 + m _2} u _1 $

Using  $m _2 >>>  m _1,$  we get $ v _1 = -u _1 $ and $ v _2 = 0 $.  

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two identical bodies moving in opposite direction with same speed, collided with each other. If the collision is perfectly elastic then

  1. after the collision both comes to rest

  2. after the collision first comes to rest and second moves in the same direction with a speed 2v

  3. after collision they recoil with same speed

  4. all the above are possible
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$mv+m(-v)=m{ v } _{ 1 }+m{ v } _{ 2 }=0\Rightarrow { v } _{ 1 }+{ v } _{ 2 }=0\$
$ \dfrac { { v } _{ 1 }-{ v } _{ 2 } }{ v-(-v) } =1\Rightarrow { v } _{ 1 }-{ v } _{ 2 }=2v\ $

$thus,\quad { v } _{ 1 }=v\ { v } _{ 2 }=-v$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

 A 6 kg mass travelling at $2.5 ms^{-1}$ collides head on with a stationary 4 kg mass. After the collision the 6 kg mass travels in its original direction with a speed of $1 ms^{-1}$. The coefficient of restitution is

  1. $1/4$
  2. $1/2$
  3. $3/4$
  4. $5/8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$e=\dfrac{v-1}{2.5}$

$=\dfrac{1.25}{2.5}=\dfrac{1}{2}$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u _0$. When the speed of the particle is $0.5 u _0$. It collides elastically with a rigid wall. After this collision.

  1. The speed of the particle when it returns to its equilibrium position is $u _0$
  2. The time at which the particle passes through the equilibrium position for the first time is $t=\pi\sqrt{\dfrac{m}{k}}$
  3. The time at which the maximum compression of the spring occurs is $t=\dfrac{4\pi}{3}\sqrt{\dfrac{m}{k}}$
  4. The time at which the particle passes through the equilibrium position for the second time is $t=\dfrac{5\pi}{3}\sqrt{\dfrac{m}{k}}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

$\cfrac{1}{2}mv _o^2=\cfrac{1}{2}Kx^2+\cfrac{1}{2}\times m(0.25)v _o^2\quad equation (1)$

After elastic collision, black speed $=v _o$
So, when, it comes back to equilibrium point its speed is $u _o$.
Amplitude, $A=\cfrac {u _o}{\sqrt{K}}$
From equation $(1)$
$x=\cfrac{\sqrt {3}u _o}{2}\sqrt {\cfrac{m}{K}}$
$\therefore t _1=\cfrac {\pi}{3\omega}=\cfrac {\pi\sqrt{m}}{3\sqrt {K}}$
Time to reach the equilibrium position for the first time $=\cfrac{2\pi}{3} \sqrt {\cfrac{m}{K}}$
Second time, it will reach at time
$=\cfrac{2\pi}{3}\sqrt {\cfrac{m}{K}}+\cfrac {T}{2}$
$=\cfrac{2\pi}{3}\sqrt{\cfrac{m}{K}}+\cfrac {2\pi\sqrt {m}}{\sqrt {K}\pi 2}$
$=\cfrac {5\pi}{3}\sqrt {\cfrac {m}{K}}$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A thin uniform rod of mass $m$ and length $l$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by(A)$\sqrt { mgl }$(B) $\sqrt { 3gl }$(c)$\sqrt { 5gl }$ (D) $\sqrt { 2gl }$  Sol. 

  1. A

  2. B

  3. C

  4. D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of energy, the potential energy of the rod (mg*l/2) is converted into rotational kinetic energy (1/2 * I * omega^2). For a rod hinged at one end, I = ml^2/3. Thus, mg*l/2 = (1/2)*(ml^2/3)*omega^2. Solving for omega gives omega = sqrt(3g/l). The velocity of the tip is v = omega*l = sqrt(3gl).

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two balls of equal mass undergo head on collision while each was moving with speed $6\ m/s$. If the coefficient of restitution is $\dfrac{1}{3}$, the speed of each ball after impact will be

  1. $18\ m/s$
  2. $2\ m/s$
  3. $6\ m/s$
  4. $4\ m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let speed of balls are $v _1$ and $v _2$.

There is no external force acting, momentum will be conserved. 
$m _1u _1+ m _2u _2= m _1v _1+m _2v _2$
$\Rightarrow m\times 6-m\times 6=mv _1+ mv _2$
$\Rightarrow v _1=-v _2$
Coefficient , $e=-\dfrac{v _1-v _2}{u _1-u _2}$  $\Rightarrow \dfrac{1}{3}= -\dfrac{v _1-v _2}{6+6}$  $\Rightarrow v _1=-2 m/s$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two particles of masses ${m} _{1}$ and ${m} _{2}$ in projectile motion have velocities ${v} _{1}$ and ${v} _{2}$ respectively at time $t=0$. They collide at time ${t} _{0}$. Their velocities become ${v'} _{1}$ and ${v'} _{2}$ at time $2{t} _{0}$ while still moving in air. The value of $\left[ \left( { m } _{ 1 }{ v' } _{ 1 }+{ m } _{ 2 }{ v' } _{ 2 } \right) -\left( { m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 } \right)  \right] $

  1. zero
  2. $({m} _{1}+{m} _{2})g{t} _{0}$
  3. $2({m} _{1}+{m} _{2})g{t} _{0}$
  4. $\cfrac{1}{2}({m} _{1}+{m} _{2})g{t} _{0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

External force = $F _{ext}=\dfrac{\Delta p}{\Delta t}=((m _{1}v _{1}^{'}+m _{2}v _{2}^{'})-(m _{1}v _{1}+m _{2}v _{2}))/2t _{0}-0$

$((m _{1}v _{1}^{'}+m _{2}v _{2}^{'})-(m _{1}v _{1}+m _{2}v _{2})=2t _{0}F _{ext}=2to(m _{1}+m _{2})g$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A free hydrogen atom in ground state is at rest. A neutron of kinetic energy $K$ collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy $2.55\ eV.$
(Assume that the hydrogen atom and neutron has same mass)

  1. Minimum value of $K$ is $25.5\ eV.$
  2. Minimum value of $K$ is $12.75\ eV.$
  3. The other photon has energy $10.2\ eV$ if $K$ is minimum.
  4. The upper energy level is of excitation energy $12.75\ eV$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To excite the atom to the n=3 level (energy -1.51 eV from -13.6 eV), the energy required is 12.09 eV. If it emits a 2.55 eV photon (n=3 to n=2), the remaining energy must be 10.2 eV (n=2 to n=1). The minimum energy K must account for the excitation and the recoil energy of the atom.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics
A cannonball is fired from a cannon at 50.0 m/s, at an angle of 30.0° above the horizontal on level ground. The goal is to determine the distance from its starting point to its ending point. The path of the cannonball is shown below. (inverse parabola).Calculate Δx.

  1. 15 cm

  2. 20 cm

  3. 30 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

In a one dimensional collision between two identical particles A and B, B is stationary and A has momentum P before impact. During impact B gives an impulse J to A. Then coefficient of restitution between the two is

  1. $\dfrac { 2J }{ P } +1$
  2. $\dfrac { 2J }{ P } -1$
  3. $\dfrac { J }{ P } +1$
  4. $\dfrac { J }{ P } -1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$p=mu$
$u=\dfrac{P}{m}$
$p-J$
$=V _A=\dfrac{P-J}{m}$     $V _B=\dfrac{J}{m}$

$\therefore \dfrac{v _{sep}}{V _{app}=\dfrac{V _B-V _A}{u}}$

$=\dfrac{\left(\dfrac{J}{m}\right)-\left(\dfrac{P-J}{m}\right)}{\dfrac{P}{m}}$

$=\dfrac{2J-P}{P}$

$=\dfrac{2J}{P}-1$