A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is and the mass of the block is Ignore the mass of the spring. [2018]

Question

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 {Nm}^{- 1}$ and the mass of the block is $2.0 kg .$ Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $1.0 kg$ moving with a speed of $2.0 {ms}^{- 1}$ collides elastically with the first block. The collision is such that the $2.0 kg$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ________. [2018]

Smart Achievers · Accepted Answer

(2.09)

Let velocities of $1 kg$ and $2 kg$ blocks just after collision be $2.0 kg .$ and $v_{2}$ respectively.

Just after collision

From momentum conservation principle,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$0 + 1 \times 2 = - 1 v_{1} + 2 v_{2} ..... (i)$

Collision is elastic. Hence $e = 1$

$e = 1 = \frac{2 - 0}{v_{1} + v_{2}}$

$\Rightarrow v_{2} + v_{1} = 2 ..... (ii)$

From eqs. (i) and (ii),

$v_{2} = \frac{4}{3} m/s, v_{1} = \frac{2}{3} m/s$

$2 kg$ block will perform SHM after collision, so spring returns to its unstretched position for the first time after,

$t = \frac{T}{2} = π \sqrt{\frac{m}{k}} = π \sqrt{\frac{2}{2}} = π = 3.14 s$

Distance or required separation between the blocks

$= | v_{1} | t = \frac{2}{3} \times 3.14 = 2.093 = 2.09 m$

Solution

Want Us to Email you About Special Offers & Updates?