Question

A block of mass 2.0 kg moving 2.0 m/s collides head on with another block of equal mass kept at rest. (a) Find the maximum possible loss in kinetic energy due to the collision. (b) If he actual loss in kinetic energy is half of this maximum, find the coefficient of restitution. 

Answer 1

It is given that:
Mass of first block, m₁ = 2 kg
Initial speed,v₁ = 2.0 m/s 
Mass of second block, m₂ = 2 kg
Initial speed of this block = 0

For maximum possible loss in kinetic energy, we assume that the collision is elastic and both the blocks move with same final velocity v (say).

On applying the law of conservation of linear momentum, we get:

m₁v₁ + m₂ ×0 = (m₁+m₂)v
2 × 2 = (2 + 2)v
⇒ v = 1 m/s

Loss in K.E. in elastic collision is give by,

\[= \frac{1}{2} m_1 v_1^2 - \frac{1}{2}\left( m_1 + m_2 \right)v\]

\[ = \left( \frac{1}{2} \right) \times 2 \times 2^2 - \left( \frac{1}{2} \right)(2 + 2) \times (1 )^2 \]

\[ = 4 - 2 = 2 J\]

\[(b) \text{ The actual loss in K . E . } = \frac{\text{ Maximum loss in K . E .}}{2} = 1 J\]
Let the final velocities of  the blocks be v₁ and v₂ respectively. The coefficient of restitution is e.
\[\therefore\] The loss in K.E. is given by, 

\[\left( \frac{1}{2} \right) \times 2 \times (2 )^2 - \left( \frac{1}{2} \right)2 \times v_1^2 - \left( \frac{1}{2} \right) \times 2 v_2^2 = 1\]

\[ \Rightarrow 4 - \left( v_1^2 + v_2^2 \right) = 1\]

\[ \Rightarrow 4 - \frac{\left( 1 + e^2 \right) \times 4}{2} = 1\]

\[ \Rightarrow 2\left( 1 + e^2 \right) = 3\]

\[ \Rightarrow 1 + e^2 = \frac{3}{2}\]

\[ \Rightarrow e^2 = \frac{1}{2}\]

\[ \Rightarrow e = \frac{1}{\sqrt{2}}\]