Test: Collisions - NEET MCQ

By conservation of momentum we get the speed of the bigger part let say, v = 1 x90 / 3
Hence we get v = 30
Thus the total KE of the system after collision is ½ (3 X 900 + 1 X 8100)
Thus KE = ½ (10800) = 5400
Now  if we apply WET to the system, as no external force has acted upon it, we get
W = ΔKE
= 5400 - 0
= 5.4 kJ

Since collision is elastic and mass is same then after velocity exchange A body will stop and B will start moving with A's velocity in the same direction.

Momentum is conserved in all types of collision whether it is elastic or inelastic where as kinetic energy is lost in sound energy in the absence of external force in inelastic collision.

The angular momentum of the two pluck will remain conservated since no extra force is involved and elastically white hockey pluck collide with stationary red hockey pluck as its a frictionless horizontal surface pluck will keep moving forward with same speed

So as no external force acts upon the system, we can say that momentum of the system remains conserved but as an internal force acts that explodes the bomb, the KE of the system is never conserved. But every fragment has some non zero momentum while the bomb initially has zero momentum.

As the balls collide head-on and no other external force is acting, momentum of the system realins conserve and so will be the energy of the system (i.e. KE)
Thus we get for the mass of other ball be m and initial velocity of mass 1 be v
1 x v = 1 x -v/3 + m x V         -1
For some V that is velocity of m after collision.
Also  ½ 1 x v² =  ½ 1 x v² /9 +  ½ m x V2                -2
Now from eq 1 we get mV = 4v/3
And from eq 2 we get  mV²  = 8v² /9
Thus we get V =  (8v² /9) / (4v/3)
=  2v/3
Thus we get m x 2v/3 = 4v/3
So m = 2kg

Before the impact the KE was ½ x m x (2g x 20) = 20mg
And let say v be the velocity after impact and for height h, v2= 2gh
Thus KE = ½ mv2 = ½m2gh = ⅗ x 20mg
Thus we get mgh = 12mg
thus h = 12 m

Concept:

1. Momentum: momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
2. The unit of momentum (P) is kg m/s.
3. Dimension: [MLT-1]
4. Law of conservation of Momentum: A conservation law stating that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
5. P₁ = P₂
6. m₁ v₁ = m₂ v₂
7. Where, P₁ = initial momentum of system, P₂ = final momentum of system, m₁ = mass of first object, v₁ = velocity of first object, m₂ = mass of second object and v₂ = velocity of second object.

Calculation:
Given:  m₁ = m kg,  m₂ = m kg,  u₁ = v m/s,  u₂ ​=  -v m/s

Let the common velocity of the combined body be V m/s

Mass of combined body      M = m + m = 2m

Applying conservation of momentum:          

m₁ v₁ + m₂ v₂ = M V

mv + (-mv) = 2mV

0 = 2mV

V = 0 m/s
Hence the correct answer will be zero (0) m/s.

Here, a sphere of mass m moving with a constant velocity v hits another stationary sphere of the same mass. As per the given problem we can write