A 10 kg ball moving with speed 2 m/s collides head on with another stationary 5 kg ball. If the collision is perfectly inelastic, then the loss in kinetic energy is -

The correct answer is option 3) i.e. 6.5 J

CONCEPT:

• Collisions in one dimension:

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

From the principle of momentum conservation,

⇒ m1v1i + m2v2i = (m1 + m2)vf   where v1i is the initial velocity of the moving object and vf is the final combined velocity after collision.

\(⇒ v_f = \frac{m_1v_{1i} +m_2v_{2i}}{m_1 +m_2}\)

The loss in kinetic energy on collision is, 

\(⇒ ΔKE = \frac{1}{2}m_1v_{1i}^2 -\frac{1}{2}(m_1+m_2)v_f^2\)

CALCULATION:

Given that:

m₁ = 10 kg, m₂ = 5 kg

u₁ = 2 m/s, u₂ = 0 m/s

Using, \( v_f = \frac{m_1u_1+m_2u_2}{m_1 +m_2}\)

\( ⇒ v_f = \frac{10(2) + 5(0)}{10+5}\)

⇒ v_f = 1.34 m/s

Loss in kinetic energy, \( ΔKE = \frac{1}{2}m_1u_1^2 -\frac{1}{2}(m_1+m_2)v_f^2\)

\( ⇒ ΔKE = \frac{1}{2}\times 10 \times 2^2 -\frac{1}{2}\times (10+5)\times1.34^2\)

⇒ ΔKE = 6.5 J