Test: Center of Mass & its motion (14 July) - JEE MCQ

The Center of Mass of a body can lie outside the body itself in cases where the body is not a solid object, such as hollow or ring-shaped objects like hoops or boomerangs. This is because the Center of Mass depends on the distribution of mass, and in such shapes, it can be located at a point in space not occupied by the material of the body.

F_A is the force on particle A

F_A = m_Aa_A = m_Av/t

F_B is the force on particle B

F_B = m_Ba_B = m_B2v/t

Since F_A= F_B

m_Av/t = m_B2v/t

So m_A = 2m_B

For the centre of mass of the system

v = (m_Av_A + m_Bv_B)/(m_A + m_B)

v = (2m_Bv – m_B2v)/(2m_B + m_B) = 0

The negative sign is used because the particles are travelling in the opposite directions

Answer: (d) Zero

Let L be the circumference of the circle

After both the particles have left from A, the particle on the left will have a velocity v and the particle on the right will have a velocity 2v.

They will first meet at point B travelling L and 2L distance.

After collision velocity will get interchanged, the body with velocity will travel with a velocity 2v now and the one having velocity 2v will travel with velocity v.

So the next collision will happen at point D travelling 2L distance and L distance.

Again the velocity will get interchanged and they will collide at point A again.

So total it will be 2 collisions before they collide again at point A.

Option (c) is the correct answer.

Just after collision

v_c= m₁v₁ + m₂v₂/m₁ + m₂

v_c= (10 x 14 + 4 x 0)/(10+4)

v_c= 10 m/s

Let the initial velocity of the bullet = V₁ m/s

The velocity with which each plate moves = V₂ m/s

Applying conservation of momentum, the initial momentum of the bullet is equal to the sum of the final momentum of the plate M1 and the momentum of the second plate including the bullet.

after piercing M

∴ mV₁ = M₁V₂ + (M₂ + m)V₂

0.02V₁ = 1 x V₂ + (2.98 + 0.02)V₂

0.02V₁ = 1 x V₂ + 3V₂

0.02V₁ = 1 x V₂ + 3V₂

V₁ = 4 V₂ /0.02

V₁ = 200V₂ ———-(1)

Let the velocity of the bullet when it comes out of the first plate = V₃

The momentum of the bullet on the first and the second plate is equal to the sum of the momentum of the second plate and the bullet.

0.02V₃ = (0.02 + 2.980 V₂) = 0.02 V₃ = 3V₂

V₃= 150V₂———-(2)

Loss percentage in the initial velocity of the bullet when it is moving between m1 and m2 is expressed as the following

Loss % ={(V₁ – V₃)/V₁ } x 100

Loss % = {(200V₂ – 150V₂ )/200V₂ } x 100

Loss % = {(200 – 150 )V₂/200V₂} x 100

Loss % = {(50 )/200} x 100

Loss % = 25%

Total kinetic energy after the collision = ½ mv₁² + ½ mv₂²= 3/2(½ mv₀²)

v₁² + v₂²= (3/2)v₀² ——–(1)

By momentum conservation

mv₀ = m(v₁+v₂) ——–(2)

(v₁+v₂)² =v₀²

v₁²+v₂²+ 2v₁v₂= v₀²

2v₁v₂ = – v₀²/2

(v₁ – v₂)² = v₁²+v₂² -2v₁v₂=(3/2)v₀² + v₀²/2

v₁ – v₂ = √2 v₀

There is an elastic collision between C and A

In an elastic collision, the velocities are exchanged if masses are the same.

∴ after the collision;

V_C = 0 V_A = v

Now the maximum compression will occur when both the masses A and B move with the same velocity.

∴ mv = (m + m) V (for system of A – B and spring)

∴ V = v/2

∴ KE of the A – B system = 1/2 x 2m (v/2)² = mv²/4

And at the time of maximum compression.

½ mv² = ½ x 2m(v/2)² + ½ kx²_max

x²_max=