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 spring of natural length 3m and spring constant 9 N/m is having one end at origin and other end attached to a block of mass 1 kg. There is a wall at x=3m. At t=0 block is released from rest at x= 1 m. Collision of block with wall is elastic. Which of the following gives position of block with time :-

  1. $x=cos\left( 3t \right) $
  2. $x=3-2sin\left( 3t+\frac { \pi }{ 2 } \right) $
  3. $x=3-\left| 2cos\left( 3t \right) \right| $
  4. $x=3-2sin\left( 3t+\frac { 3\pi }{ 2 } \right) $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The spring constant k = 9 N/m and mass m = 1 kg give an angular frequency omega = sqrt(k/m) = 3 rad/s. Natural length is 3m, block is released from x = 1m with natural length of spring 3m means initial stretch is 2m. Without the wall, it would oscillate as x = 3 - 2cos(3t). Because of the elastic wall at x = 3m, the motion on the right side is reflected, leading to the absolute value form x = 3 - abs(2cos(3t)).

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

A steel ball moving with a velocity $\overline{v}$ collides with an identical ball originally at  rest. The velocity of the first ball after the collision is :

  1. $\left(-\dfrac{1}{2}\right)\overline{v}$
  2. $-\overline{v}$
  3. $\overline{v}$

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

Here, a steel ball moving with a velocity $\bar v$ collides with an identical ball originally at  rest. hence, masses of two steel balls are equal. For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity, thus, the velocity of the first ball after the collision is zero.

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

In the elastic collision of heavy vehicle moving with a velocity 10 ms$^{-1}$ and a small stone at rest, the stone will fly away with a velocity equal to : 

  1. 40 ms$^{-1}$
  2. 20 ms$^{-1}$
  3. 10 ms$^{-1}$
  4. 5 ms$^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is 
for heavy particle, $v _1 = u _1$
for light particle, $v _2 = 2u _1 - u _2$
since, $u _2 = 0$ hence, 
$v _2 = 2u _1$
Therefore, the stone will fly away with a velocity equal to 
$v _2 = 2u _1 = 2(10) = 20 ms^{-1}$

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

A block of mass 100$\mathrm { g }$ attached to a spring of stiffness 100$\mathrm { N } / \mathrm { m }$ is lying on a frictionless floor as shown. block is moved to compress the spring by 10 cm and released. If the collision with the wall is elastic then the time period of oscillations. (in seconds) 

  1. 0.133

  2. 13.3

  3. 0.26

  4. 0.3
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two particles of masses $ {m} _{1}, {m} _{2} $ movie with initial velocities $ u _{1} \text { and } u _{2} $.On collision, one of the particles get excited to higher level, after absorbing energy If final velocities of particles be $  v _{1}  $ and $  v _{2}  $ then we must have :

  1. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}-\varepsilon
    $
  2. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}+\varepsilon=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}
    $
  3. $
    \dfrac{1}{2} m _{1}^{2} u _{1}^{2}+\dfrac{1}{2} m _{2}^{2} u _{2}^{2}-\varepsilon=\dfrac{1}{2} m _{1}^{2} v _{1}^{2}+\dfrac{1}{2} m _{2}^{2} v _{2}^{2}
    $
  4. $
    m _{1}^{2} u _{1}+m _{2}^{2} u _{2}-\varepsilon=m _{1}^{2} v _{1}+m _{2}^{2} v _{2}
    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Total\, \, initial\, \, energy\, \, of\, \, two\, \, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 } \ Total\, \, final\, \, energy\, \, of\, \, two\, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ U\sin  g\, \, energy\, \, conservation\, \, principle, \ \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }=\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ \therefore \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }-\in =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 } \end{array}$

Hence,
option $(C)$ is correct answer.

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

A moving sphere of mass m suffer a perfect elastic collision (not head on) with an  equally massive stationary sphere. after collision both fly off at angle $\theta $ value of which is :

  1. 0
  2. $\pi $

  3. indeterminate
  4. $\pi /2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an elastic collision between two equal masses where one is initially at rest, the angle between the final velocity vectors is always 90 degrees (pi/2).

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

A rubber ball is bounced on the floor of a room which has its ceiling at a height of  $3.2{ m }$  from the floor. The ball hits the floor with a speed of  $10 m / { s },$  and rebounds vertically up. If all collisions simply reverse the velocity of the ball, without changing its speed, then how long does it take the ball for a round trip, from the moment it bounces from the floor to the moment it returns back to it ? Acceleration due to gravity is  $10 m / s ^ { 2 }.$

  1. $4 s$
  2. $2 s$
  3. $0.8 s$
  4. $1.2 s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time taken to reach the highest point after the rebound at speed v = 10 m/s is t_up = v/g = 10/10 = 1 s. However, the ceiling is at a height of 3.2 m. Using kinematic equation y = v*t - (1/2)*g*t^2, at y = 3.2m we get 3.2 = 10*t - 5*t^2, which solves to t = 0.4 s (the ball hits the ceiling before reaching its natural peak of 5m). The collision with the ceiling reverses the velocity, so the round trip time until it returns to the floor is 0.8 s.

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

 A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly- elastic collision with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value of m is/are :

  1. 1 kg only

  2. 1 kg , 9kg

  3. 1 kg, 6kg

  4. 6kg only
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial kinetic energy of the 3 kg ball is (1/2)3(4)^2 = 24 J. After an elastic collision, its kinetic energy is 6 J. Using the conservation of kinetic energy and momentum for an elastic collision in 1D, the final speed of the 3 kg mass can be found, leading to possible mass values of m = 1 kg or m = 9 kg.

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

In an elastic collision the K.E of one body decreases by $100 J$. If the masses colliding bodies are in the ratio 3:4 the K.E of the other body increase by 

  1. $\dfrac{400}{3} J$
  2. $\dfrac{500}{3} J$
  3. $100 J$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In any isolated system undergoing an elastic collision, total kinetic energy is conserved. Thus, the decrease in kinetic energy of one body must equal the increase in kinetic energy of the other body. Since one body loses 100 J, the other body gains exactly 100 J, regardless of their mass ratio.