Centre Of Mass NAT Level - 2 - IIT JAM MCQ

The initial extension in spring is 
Just after collision of B with A the speed of combined mass is  v/2.

For the spring to just attain natural length of combined mass must rise up by (see figure.) and comes to rest.

Applying conservation of energy between initial and final states

Alternate solution by SHM

The correct answer is: 6

Let u and v be speed of wedge A and block B at just after the block B gets off the wedge A. Applying conservation of momentum in horizontal direction, we get.

mu = mv      ...(1)

Applying conservation of energy between initial and final state as shown in figure (1), we get

solving (1) and (2) we get

   ...(3)

At the instant block B reaches maximum height h' on the wedge C(figure 2), the speed of block B and wedge C are v'.

Applying conservation of momentum in horizontal direction, we get

Applying conservation of energy between initial and final state

Solving equations (3), (4) and (5) we get

The correct answer is: 25

Maximum compression will take place when the blocks move with equal velocity. As no net external horizontal force acts on the system of the two blocks, the total linear momentum will remain constant. If v is the common speed at maximum compression, we have,

(1 kg) (2m/s) = (1kg)v + (1kg)v

or v = 1m/s

Initial kinetic energy 

Final kinetic energy =

or x = 0.2m

The correct answer is: 20

Let the three mutually perpendicular direction be along x, y and z-axis respectively.

By linear momentum conservation,

= 3E₀ + 3E₀
= 6E₀
The correct answer is: 6

Neglecting gravity

u = ejection velocity w.r.t. balloon
m₀ = initial mass
m₁ = mass at any time t.

The correct answer is: 1.386

For first collision v = 10 m/s, 

velocity of separation = e velocity of approach

v₂ – v₁ = 5m/s
for second collision

The correct answer is: 2.5

Force F on plate = force exerted by dust particles
= force on dust particles by the plate
= rate of change of momentum of dust particles
= mass of dust particles striking the plate per unit time × change in velocity of dust particles

The correct answer is: 2

For 1^st collision
Since 

e = 1
v = v₁ + v₂
⇒ v₂ = v – v₁   ...(1)
By momentum conservation :

For second collision :

By momentum conservation :

From (2) and (4) :

Therefore B can not collide with C for the second time.
Hence; total number of collision is 2.
The correct answer is: 2

Since e = 1/5

∴  Final normal component of velocity 
As the angle of rebound is equal to the angle before impact.

Therefore, both normal and tangential components of velocities must change by the same factor,
∴ Tangential velocity after impact becomes 
Let the time of impact be Δt.

where N is the normal force imparted on the ball by the wall.

The correct answer is: 5

Let v₁, v₂ and v₃  be velocities of block 1, 2 and 3 after suffering collision each.
mv = mv₁ + Mv₂
and v₁ – v₂ = –v
v₁ – v₂ = –v

The correct answer is: 4.236