Question

A massive ball moving with a speed 'u' collides with a tiny ball having a very small mass kept at rest. Then the speed of the second ball immediately after the impact will be -

1. u
2. 0.5u
3. √u
4. 2u

Answer 1

Correct Answer - Option 4 : 2u

The correct answer is option 4) i.e. 2u 

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      ----(1)

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      ----(2)

From (1) and (2),

\(\Rightarrow v_{2f} = \frac{2m_1v_{1i} + v_{2i} (m_2 - m_1)}{m_1 + m_2}\)

​CALCULATION:

• Since the smaller ball is at rest initially, v_2i = 0. Let v_2f be the speed of the smaller ball after collision.

⇒ m1v1i = (m1 + m2)vf 

\(\Rightarrow v_{2f} = \frac{2m_1v_{1i} + v_{2i} (m_2 - m_1)}{m_1 + m_2}= \frac{2m_1v_{1i} + 0 (m_2 - m_1)}{m_1 + m_2}\)

\(\Rightarrow v_{2f} =\frac{2m_1v_{1i} }{m_1 + m_2}\)

• Since the mass of the smaller ball is very less, m₂ << m₁. Therefore, m₁ + m₂ ≈ m₁

\(\Rightarrow v_{2f} =\frac{2m_1v_{1i} }{m_1 } = 2 v_{1i}\)

• Given that, v_1i = u. Thus, v_2f = 2u