Tag: collisions in one dimension

Questions Related to collisions in one dimension

For head-on collision between two colliding balls of equal radii $r$, the impact parameter is equal to

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

  1. $2r$

  2. $Zero$

  3. $More \ than \ 2r$

  4. $Less\ than \ 2r$

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Correct Option: A

In a one-dimensional collision between two particles, their relative velocity is $\bar{v _1}$ before the collision and $\bar{v _2}$ and the collision.

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

  1. $\bar{v _1} = \bar{v _2}$ if the collision is elastic.

  2. $\bar{v _1} = - \bar{v _2}$ if the collision is elastic.

  3. $|\bar{v _2}| = |\bar{v _1}|$ in all cases.

  4. $\bar{v _1} = - k \bar{v _2}$ in all cases, where k $\geq$ 1.

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Correct Option: B,C,D
Explanation:

If $ { v } _{ 1 }$ is relative velocity before collision.
if ${ v } _{ 2 } $ is relative velocity before collision.
$e\le 1\$
$ e=\dfrac { { v } _{ 1 } }{ { -v } _{ 2 } } $


so $\left| { v } _{ 1 } \right| \ge \left| { v } _{ 2 } \right| $
also due to impact the ratios of velocity get changed,relative velocities D is correct. also, for elastic collision e$=$1.
So, option B is correct only if both particles have equal masses,not in general.

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 impulse $J$ to $A$.

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

  1. The total momentum of the '$A\ plus\ B$' system is $p$ before and after the impact, and $(p - J)$ during the impact.

  2. During the impact, $A$ gives impulse $J$ to $B$.

  3. The coefficient of restitution is $\displaystyle \dfrac{2 J}{p} - 1$

  4. The coefficient of restitution is $\displaystyle \dfrac{ J}{p} + 1$

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Correct Option: B,C
Explanation:

Let  $u=$ speed of A before impact. Thus,  $p=mu$.
Let $v _1, v _2 = $ speeds of  $A$ and $B$ after impact.
$u = v _1 + v _2 $ and $v _1 - v _2 = - eu$
$u = v _1 + v _2$ and $v _1 - v _2 = - eu$


$\therefore v _1 = \dfrac{1}{2} u (1-e)$ and $v _2 = \dfrac{1}{2} u (1 + e)$

$J = mv _2 = m \displaystyle \left [ \dfrac{1}{2} u (1 + e) \right ] = \dfrac{1}{2} p (1 + e)$

$\Rightarrow e=\dfrac{2J}{p}-1$

A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after collision will be, if the co-efficient of restitution is e:

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

  1. $\displaystyle \frac{1 - e}{1 + e}$

  2. $\displaystyle \frac{e - 1}{e + 1}$

  3. $\displaystyle \frac{1 + e}{1 - e}$

  4. $\displaystyle \frac{e + 1}{e - 1}$

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Correct Option: A
Explanation:

The law of conservation  of linear momentum tells us that the overall momentum before the collision must be equal to the overall momentum after a collision.

Since the spheres have identical masses, we can write

$mu + m\times 0 = mv _A + mv _B$

$u = v _A+v _B$

From the definition of the coefficient of restitution, we know that

$e = \dfrac{v _B - v _A}{u}$

solving above two equations

$e \times ( v _A+v _B) = v _B - v _A$

$v _B(1-e) = v _A (1+e)$

$\dfrac{v _A}{v _B} = \dfrac{1-e}{1+e}$

In head on elastic collision of two bodies of equal masses:

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

  1. the velocities are interchanged

  2. the speeds are interchanged

  3. the momentum are interchanged

  4. the faster body slows down and the slower body speeds up

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Correct Option: A,B,C,D
Explanation:

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. Hence, the velocities are interchanged i.e. the speeds are interchanged which in turn interchanges the momentum. Also, if target have some velocity then the faster body slows down and the slower body speed up.

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 :-

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

  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) $

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Correct Option: A

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 :

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

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

  2. $-\overline{v}$

  3. $\overline{v}$

  4. zero

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Correct Option: D
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.

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 : 

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

  1. 40 ms$^{-1}$

  2. 20 ms$^{-1}$

  3. 10 ms$^{-1}$

  4. 5 ms$^{-1}$

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Correct Option: B
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 
$v _2 = 2u _1 = 2(10) = 20 ms^{-1}$

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) 

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

  1. 0.133

  2. 13.3

  3. 0.26

  4. 0.3

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Correct Option: A

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 :

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

  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}
    $

Show answer
Correct Option: C
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.