Test: Collisions (June 24) - JEE 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.

1. In an elastic collision, both kinetic energy and linear momentum are conserved

2. The white puck is moving and collides with the stationary red puck. Therefore, the kinetic energy of the white puck before the collision is not necessarily conserved individually, but the total kinetic energy of the system (both pucks) is conserved

3. The linear momentum of the white puck alone is not conserved because it is involved in the collision. However, the total linear momentum of the two-puck system is conserved since there are no external forces acting on it

4. Based on the analysis:

   • Statement (1) is false because the kinetic energy of the white puck alone is not conserved
   • Statement (2) is false because the linear momentum of the white puck alone is not conserved
   • Statement (3) is true because the linear momentum of the two-puck system is conserved

Thus, the only statement that must be true is statement (3)

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

Correct Answer :- a

Explanation :Here, a body of mass m moving with a constant velocity v hits another body of the mass m moving with same velocity v but in the opposite direction, and sticks to it. We can write
m₁​u₁​−m₂​u₂​=(m_1​+m₂​)v
As m₁​=m₂​=m and u₁​=u₂​=v
2mv=0
v=0

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