Question

A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of u. 

Answer 1

It is given that:
Mass of particles = 100 g 
Initial speed of the first particle = u 
Final K.E. of the system after collision = 0.2J

Initial K.E. of the system, before collision = \[\frac{1}{2}m u^2 + 0 \]

i.e. Initial K.E. = \[\frac{1}{2} \times 0 . 1 \times u^2 = 0 . 05 u^2\]
Let  v₁ and v₂ be the final velocities of the first and second block respectively.    
Let  v₁ and v₂ be the final velocities of the first and second block respectively.    
By law of conservation of momentum, we know: 
\[m v_1 + m v_2 = mu\]

\[\Rightarrow v_1 + v_2 = u . . . (1)\]

\[( v_1 - v_2 ) + e( u_1 - u_2 ) = 0\]

\[ \Rightarrow eu = v_2 - v_1 . . . (2) [\text{ Putting } u_2 = 0, u_1 = u]\]

\[\text{  Adding the equations (1) and (2), we get:}\]

\[ 2 v_2 = (1 + e)u\]

\[ \Rightarrow v_2 = \left( \frac{u}{2} \right)(1 + e)\]

\[ \therefore v_1 = u - \frac{u}{2}(1 + e)\]

\[ v_1 = \frac{u}{2}(1 - e)\]

\[\text{ According to given condition, }\]

\[\frac{1}{2}m v_1^2 + \frac{1}{2}m v_2^2 = 0 . 2\]

\[ \Rightarrow v_1^2 + v_2^2 = 4\]

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

\[ \Rightarrow u^2 = \frac{8}{1 + e^2}\]
For maximum value of u, denominator should be minimum in the above equation.
i.e. e = 0
⇒ u² = 8
\[\Rightarrow u = 2\sqrt{2} \text{ m/s}\]
For minimum value of u, denominator should have maximum value.
i.e. e = 1
⇒ u² = 4
⇒ u = 2 m/s .