Test: MCQs (One or More Correct Option): Momentum and Impulse | JEE Advanced - JEE MCQ

If we consider the two particles as a system then the external force acting on the system is the gravitational pull (m₁ + m₂) g.

Just after collision

Note : Spring force is an internal force, it cannot change the linear momentum of the (two mass + spring) system.
Therefore v_c remains the same.

For vertical motion of bullet or ball
u = 0,  s = 5m,  t = ? ,  a = 10m/s²

For horizontal motion of ball

For horizontal motion of bullet

Applying conservation of linear momentum during collision, w e get

Activity B to M for particle thrown upwards

Applying conservation of linear momentum in Y-direction
2mv sinθ = mv1 = mu₀cosα ...(ii)    [from (i)]
Applying conservation of linear momentum in X-direction
2mv cosθ = mu₀ cosα ...(iii)
on dividing (ii) and (iii) we get

(a) is wrong because the momentum of ball  changes in magnitude as well as direction.
(b) is wrong because on collision, some mechanical energy is converted into heat, sound energy.

(c) is correct because for earth + ball system the impact force is an internal force.
(d) is correct.

As one piece retraces its path, the speed of this piece just after explosion should be v cos θ

(At highest point just after explosion) NOTE THIS STEP
Applying conservation of linear momentum at the highest point;

In situation 1, mass C is moving towards right with velocity v. A and B are at rest.
In situation 2, which is just after the collision of C with A, C stop and A acquires a velocity v. [head-on elastic collision between identical masses]
When A starts moving towards right, the spring suffer a compression due to which B also starts moving towards right. The compression of the spring continues till there is relative velocity between A  and B. When this relative velocity becomes zero, both A and B move with the same velocity v' and the spring is in a state of maximum compression.
Applying momentum conservation in situation 1 and 3,

∴ K.E. of the system in situation 3 is

This is the kinetic energy possessed by A – B system (since, C is at rest).
Let x be the maximum compression of the spring.
Applying energy conservation

KEY CONCEPT Use law of conservation of linear momentum.
The initial linear momentum of the system is  Therefore the final linear momentum should also be zero.

Option a :
 Final momentum.
It is given that a₁, b₁, c₁, a₂, b₂ and c₂ have non-zero values. If a₁ = x and a₂ = – x. Also if b₁ = y and b₂ = – y then the  components become zero. But the third term having  component is non-zero. This gives a definite final momentum to the system which violates conservation of linear momentum, so this is an incorrect option.

Option d:

because b₁ ≠ 0
Following the same reasoning as above this option is also ruled ou

According to law of conservation of linear momentum 1 × u₁ + 5 × 0 = 1 (–2) + 5 (v₂)
⇒ u₁ = – 2 + 5 v₂ ...(i)

The coefficient of restituition

On solving (i) & (ii) we get desired results.

The particle collides elastically with rigid wall. Here 

i.e. the particle rebounds with the same speed. Therefore the particle will return to its equilibrium position with speed u₀. option (a) is correct.
The velocity of the particle becomes 0.5u₀ after time t.
Then using the equation V = V_max cos wt we get 0.5u₀ = u₀ cos wt

The time period 
The time taken by the particle to pass through the equilibrium for the first time   Therefore option (b) is incorrect
The time taken for the maximum compression

  Therefore option c is incorrect.
The time taken for particle to pass through the equilibrium position second time

option (d) is correct.