Tag: collisions

Questions Related to collisions

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] $

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

  1. zero

  2. $({m} _{1}+{m} _{2})g{t} _{0}$

  3. $2({m} _{1}+{m} _{2})g{t} _{0}$

  4. $\cfrac{1}{2}({m} _{1}+{m} _{2})g{t} _{0}$

Show answer
Correct Option: C
Explanation:

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

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

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)

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

  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 $12.75\ eV$.

Show answer
Correct Option: C
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.
collisions in one dimension collisions work, energy and power mechanics physics

  1. 15 cm

  2. 20 cm 

  3. 30 cm

  4. 10 cm

Show answer
Correct Option: D

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

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

  1. $\dfrac { 2J }{ P } +1$

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

  3. $\dfrac { J }{ P } +1$

  4. $\dfrac { J }{ P } -1$

Show answer
Correct Option: B
Explanation:
$p=mu$
$u=\dfrac{P}{m}$
$p-J$
$=V _A=\dfrac{P-J}{m}$     $V _B=\dfrac{J}{m}$

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

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

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

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

A gun fires a shell of mass $1.5 $kg with velocity of $150$ m/s and recoils with a velocity of $2.5$ m/s. 

Calculate the mass of the gun.

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

  1. $20$ kg

  2. $30$ kg

  3. $90$ kg

  4. $60$ kg

Show answer
Correct Option: C
Explanation:

Mass of shell $=1.5$kg

Velocity of shell ${V _1} = 159\,m/s$
Recoil velocity $=2.5$ m/s
mass of gun $=M$
According to the conservation of momentum,
$m{v _1} = m{v _2}$
$\begin{array}{l} 1.5\times 150=M\times 2.5 \ M=\dfrac { { 1.5\times 159 } }{ { 2.5 } }  \ M=90\, kg \end{array}$

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 walk. After this collision:

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

  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 througout the equilibrium position for the second time is $t= \dfrac{4\pi}{3}\sqrt{\dfrac{m}{k}}$

Show answer
Correct Option: A

A ball hits the floor and rebounds after an elastic collision. In this case:

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

  1. the momentum of the ball just after the collision is same as that just before the collision.

  2. the kinetic energy of the ball remains same during collision.

  3. the total momentum of the ball and the earth is conserved.

  4. the total energy of the ball and the earth remains the same.

Show answer
Correct Option: C,D
Explanation:

Since the velocity before and after the collision change hence momentum of the ball will change. So (a) is not true.  


Since the collision is inelastic a part of the mechanical energy is lost hence (b) is not true.  

Taking earth and the ball as a system there is no external force on the system. Hence the total momentum of the ball and the earth is conserved. So (c) is true.  

From the conservation principle of the energy, the total energy of the ball and the earth remains the same. Hence (d) is true.

Two identical spheres move in opposite direction with speed $v _1$ and $v _2$ and pass behind an opaque screen, where they may either cross without touching ( Event 1) or make an elastic head-on collision ( Event 2)

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

  1. We can never make out which events has occured

  2. We cannot make out which event has occured only if $v _1= v _2$

  3. We can always make out which event has occured

  4. We can make out which event has occured only if

Show answer
Correct Option: A

A bullet of mass m is fired into a large block of wood of mass M with velocity v. The final velocity of the system is?

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

  1. $\left(\dfrac{m}{M-m}\right)v$

  2. $\left(\dfrac{m+M}{M}\right)v$

  3. $\left(\dfrac{M-m}{M}\right)v$

  4. $\left(\dfrac{m}{m+M}\right)v$

Show answer
Correct Option: D
Explanation:

u is initial velocity 

v is finall velocity 
m1 is m 
m2 is M 
u1 is v 
u2 is 0 
${ v } _{ 1 }={ v } _{ 2 }={ v } _{ 3 }$
$total\quad initial\quad momentum=finall\quad momentum$
${ m } _{ 1 }{ u } _{ 1 }+{ m } _{ 2 }{ u } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$
$mv+M\times 0=m{ v } _{ 3 }+{ Mv } _{ 3 }$
$mv=\left( m+M \right) { v } _{ 3 }$
${ v } _{ 3 }=\dfrac { mv\quad  }{ m+M } $

sphere collides with another sphere of identical mass kept at rest. Mier collision, the two spheres move. The collision is perfectly inelastic, then the angle between the directions of motion of the two spheres is

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

  1. ${ 0 }^{ o }$

  2. ${ 45 }^{ o }$

  3. different from ${ 90 }^{ o }$

  4. ${ 90 }^{ o }$

Show answer
Correct Option: A