Tag: collisions

Questions Related to collisions

A body of mass 2kg is projected upward from the surface of the ground at t$=$0 with a velocity of 20 m/s. One second later a body B, also of mass 2 kg, is dropped from a height of 20 m. If they collide elastically, then velocities just after collision are 

  1. V$ _A = $ -5 m/s downward , V$ _B = $ 5 m/s upward

  2. V$ _A = $ 10 m/s downward , V$ _B = $ 5 m/s upward

  3. V$ _A = $ 10 m/s upward , V$ _B = $ 10 m/s downward

  4. both move downward with velocity 5 m/s


Correct Option: A

In an inelastic collision-

  1. Momentum of the system is always conserved.

  2. Velocity of separation is less than the velocity of approach.

  3. The coefficient of restitution can be zero.

  4. All of the above.


Correct Option: D
Explanation:

In inelastic collision momentum of the system is always conserved if $F _{ext}=0$.
Velocity of separation is less than the velocity of approach since co-efficient of restitution e < 1
$e=0$ for a perfectly inelastic collision since the colliding particles stick together after collision.
Hence, option (D) is correct.

A mass $m$ moves with velocity $v$ and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity $\frac { v }{ \sqrt { 3 }  } $ in a direction perpendicualr to the initial direction of motion. find the speed of the second mass after collision.

  1. $v$

  2. $\sqrt { 3v } $

  3. $\frac { 2 }{ \sqrt { 3 } } v$

  4. $\frac { v }{ \sqrt { 3 } } $


Correct Option: C

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed v. What is the mass of other body and the speed of the center of mass of two bodies ?

  1. $ 1.0 kg and \frac {2}{3}v $

  2. $ 1.2 kg and \frac{5}{8} $

  3. $ 1.4 kg and \frac {10}{17} v $

  4. $ 1.5 kg and \frac {4}{7} v $


Correct Option: B
Explanation:
$m _{1}=2\ kg$
$u _{1}=4\ m/s$
$v _{1}=\dfrac {u _{1}}{4}=\dfrac {4}{4}=1\ m/s$
$u _{2}=0$
since nothing is given so we would take it as elastic collision. 
hence conservation of momentum is given by:
$m _{1}u _{1}+m _{2}u _{2}=m _{1}v _{1}+m _{2}v _{2}$
$2\times 4+0=2\times 1+m _{2}v _{2}$
$m _{2}v _{2}=6$
according to the conservation kinetic energy:
$\dfrac {1}{2}m _{1}u _{1}^{2}+\dfrac {1}{2}m _{2}u _{2}^{2}=\dfrac {1}{2}m _{1}v _{1}^{2}+\dfrac {1}{2}m _{2}v _{2}^{2}$
$2\times 16+0=2\times 1+m _{2}v _{2}^{2}$
$m _{2}v _{2}^{2}=30$
$\dfrac {m _{2}v _{2}^{2}}{m _{2}v _{2}}=\dfrac {30}{6}$
$v _{2}=5\ m/s$
$m _{2}=1.2\ kg$


Two objects of mass 3 kg and 2 kg move along x and y ais with $ 4 m/s^2 $ and $ 3 m/s ^2 $ respectively on a horizontal smooth table.after collision the bodies stick together.then :

  1. Heat generated in the process is 15 joules

  2. Heat generated in the process is 18 joules

  3. direction of motion x-axis after collision is $ 60^0 $

  4. direction of x-axis after collision is $ tan{-1} \left( \frac { 1 }{ 3 } \right) $


Correct Option: B

$n$ balls each of mass $m$ impinge elastically each second on a surface with velocity $u$. The average force experienced by the surface will be

  1. $mnu$

  2. $2\ mnu$

  3. $4\ mnu$

  4. $mnu/2$


Correct Option: B
Explanation:
  • Elastically means the balls get bounced with $same$ speed i.e $u$
  • so the change in momentum will be $P _2-P _1=-mu-mu=-2mu$
  • so the momentum tranferred to the surface will be $-(-2mu)=2mu$
  • thus the total momentum transferred by $n$ balls will be $2mnu$
  • force $F=\dfrac{Momentum}{time}=\dfrac{2mnu}{1sec}=2mnu$

A body moving towards a finite body at rest collides with it. It is possible that :

  1. both the bodies come to rest

  2. both the bodies move after collision

  3. the moving body comes to rest and stationary body starts moving

  4. the stationary body remain stationary and moving body changes its direction


Correct Option: B,C
Explanation:

If a body moving towards a finite body at rest collides with it then momentum will be conserved and hence, the velocities after collision are may be in inverse proportion to their masses or may get interchanged. Hence, both the bodies move after collision or the moving body comes to rest and stationary body starts moving.

A body moving towards a finite body at rest collides with it. It is possible that:

  1. both the bodies come to rest

  2. both the bodies moves after collision

  3. the moving body comes to rest and the stationary body starts moving

  4. the stationary body remains stationary, the moving body changes its velocity


Correct Option: B,C
Explanation:

both bodies cannot come to rest as  it will violate the law of conservation of momentum.
Option (D) will also violate conservation of momentum.

A mass $m _1$ moves with a great velocity. It strikes another mass $m _2$ at rest in a head on collision and comes back along its path with a low speed after collision. Then :

  1. $m _1 > m _2$

  2. $m _1 = m _2$

  3. $m _1 < m _2$

  4. there is relation between $m _1$ and $m _2$


Correct Option: C
Explanation:

In a head-on elastic collision between a small projectile and a more massive target, the projectile will bounce back with low speed and the massive target will be given a very small velocity. Hence, $m _1 < m _2$

A body moving towards a body of finite mass at rest collides with it. It is possible that :

  1. both bodies come to rest

  2. the stationary body remains stationary and the moving body rebounds.

  3. the moving body stop and the body at rest start moving.

  4. all of the above are correct.


Correct Option: C
Explanation:

From the very fundamental law of collision we know that in a collision momentum of the system is conserved .
Therefore the only possible option is C.