Question

A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly elastic collision with a stationary ball. After the impact is over, the kinetic energy of the 3 kg ball is 9 J. Then the mass of the stationary ball is -

1. 1 kg
2. 0.72 kg
3. 0.56 kg
4. 0.33 kg

Answer 1

Correct Answer - Option 2 : 0.72 kg

The correct answer is option 2) i.e. 0.72 kg

CONCEPT:

• Elastic collision: Elastic collision is a phenomenon where the collision of objects takes place such that the total linear momentum and kinetic energy of the system are conserved.
• Collisions in one dimension:

​Let m1 and m2 be the masses of two objects that undergo elastic collision. 

From the principle of momentum conservation,

⇒ m1v1i + m2v2i = m1v1f + m2v2f      

where m1, m2 are the masses of the colliding bodies, v1i, v2i are the initial velocity and v1f and v2f are their final velocities.

From the principle of kinetic energy conservation,

⇒ m1v1i2 + m2v2i2 = m1v1f2 + m2v2f2      

​EXPLANATION:

Given that:

m₁ = 3 kg, u₁ = 4 m/s, KE₁ = 9 J

Let the mass and velocities of the second ball be m₂ and u₂, v₂ respectively. 

After the impact, kinetic energy of the first ball, \(KE_1 = \frac{1}{2}m_1 v_1 ^2 = 9 \: J\)

\(⇒ \frac{1}{2} × 3 × v_1 ^2 = 9\)

⇒ v₁ = √6 m/s     

From the principle of momentum conservation,

⇒ m1u₁ + m2u₂ = m1v1 + m2v2

⇒ m1u1 = m1v1 + m2v2    (As the second ball is stationary, u₂ = 0 m/s)

⇒ 3 × 4 = 3 × √6 + m2v2   

⇒  m₂v₂ = 4.65     ----(1)

From the principle of kinetic energy conservation,

⇒ m1u1² = m1v1² + m2v2²

⇒ 3 × 4² = 3 × √62 + m2v22

⇒ m₂v₂² = 30      ----(2)

Substituting (1) in (2),

⇒ 4.65 v₂ = 30

⇒  v₂ = 6.45 m/s   ----(3)

Substituting (3) in (1),

⇒ m₂ (6.45) = 4.65

⇒  m₂ = 0.72 kg