91影视

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 .