“Relative” is an important word.In Fig. 9-72, block Lof mass${\mathit{m}}_{\mathbf{L}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{kg}$and block Rof mass ${\mathit{m}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{kg}$are held in place witha compressed spring between them.When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block La release speed of 1.20 m/srelativeto the floor, how far does block R travel in the next 0.800 s? (b) If, instead, the spring gives block La release speed ofrelativeto the velocity that the spring gives block R, how far does block Rtravel in the next 0.800 s?

Mass of the block L is,$m_L=1.00\mathrm{kg}$ .

Mass of the block R is,$m_R=0.500\mathrm{kg}$ .

Speed of the block L relative to the floor is, $v_L=-1.20\mathrm{m}/\mathrm{s}$.

The speed of block L relative to the speed of the block R is,
$v_r=-1.20\frac{\mathrm{m}}{\mathrm{s}}$.

The time is given as, t = 0.800 s.

Using the law of conservation of momentum, we can find the velocity of block R at 0.800 s after release if the spring gives block L a release speed of 1.20 m/s relative to the floor. From this, we can easily find the distance traveled by block R corresponding to it. Similarly, we can find the distance traveled by the block R in 0.800 s after release if the spring gives block L a release speed of 1.20 m/s relative to the velocity of R using the concept of relativity.

Formula:

The momentum of the system before release = Momentum of the system after release.

x = v t

The total momentum of the system is conserved if no external force acts on it.

According to the law of conservation of momentum,

The momentum of the system before release = Momentum of the system after release

$0=\left(m_L{v}_L+m_R{v}_R\right)$ (1)

Here is the velocity of block R.

Substitute the values in the above expression, and we get,

$$\begin{gathered} 0=\left(\left(1.00\right)\left(-1.20\right)+\left(0.500\right)\left(v_R\right)\right) \\ {v}_R=2.40 \end{gathered}$$

The traveled distance can be calculated as,

$x_R=v_{R}t$

Substitute the values in the above expression, and we get,

$$\begin{gathered} x_R=\left(2.4\right)\left(0.800\right) \\ =1.92\mathrm{m} \end{gathered}$$

Therefore, the distance traveled by the block R in 0.800 s after release is 1.92 m.

Since,

The velocity of block L can be written as,

$v_L=v_R+v_r$

Substitute this value in equation 1, and we get,

$0=\left(m_L\left(v_R+v_r\right)+m_R{v}_R\right)$

Substitute the values in the above expression, and we get,

$$\begin{gathered} 0=\left(\left(1.00\right)\left(v_R-1.20\right)+\left(0.500\right)\left(v_R\right)\right) \\ 1.5v_R=1.2 \\ {v}_R=0.8\mathrm{m}/\mathrm{s} \end{gathered}$$

The traveled distance can be calculated as,

$x_R=v_{R}t$

Substitute the values in the above expression, and we get,

$$\begin{gathered} x_R=\left(0.8\right)\left(0.800\right) \\ =0.640\mathrm{m} \end{gathered}$$

Therefore, the distance traveled by the block R in 0.800 s after release is 0.640 m .