Figure shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6.00 kg) and blocks L and R (each ofmas m = 2.00 kg) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence: (1) At time t = 0, block L is shot to the left with a speed of 3.00 m/srelative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.80 s, block R is shot to the right with a speed of 3.00 m/srelative to the velocity that block C then has. At t = 2.80 s, (a) What are the velocity of block C and (b) What are the positions of its center?

The mass of the block C is,$M_c=6.00\mathrm{kg}$.

The mass of the block L is,$m_L=2.00\mathrm{kg}$.

The mass of the block R is, $m_R=2.00\mathrm{kg}$.

At time t = 0 , speed of the block L is, $v_L=3.00\mathrm{m}/\mathrm{s}$.

At time t = 0.80 s, speed of the block R is, ${\mathrm{v}}_{\mathrm{R}}=3.00\mathrm{m}/\mathrm{s}$.

By using the conservation of momentum, find the velocity ${\mathbf{v}}_{\mathbf{2}}$of block C. Also, by using the values $\mathbf{}{\mathbf{v}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{2}}\mathbf{}$, find the position of its center. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$m_L\left(v_1-v_L\right)+\left(m_c+m_R{v}_1=0\right)$

$x=v∆t$

Where, $m_L,m_c,m_R$are masses of block L, C, and R respectively. And v,$v_1$are velocities,x is displacement, and$∆t$ is time period.

The velocity of block L is$\left(v_1-3\right)\hat{i}$. Thus, momentum conservation gives,

$$\begin{gathered} m_L\left(v_1-v_L\right)+\left(M_c+m_R\right)v_1=0 \\ {m}_L{v}_1+M_c{v}_1+m_R{v}_1=m_L{v}_L \\ \left(m_L+M_c+m_R\right)v_1=m_L{v}_L \\ {v}_1=\frac{m_L{v}_L}{m_L+M_c+m_R} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} v_1=\frac{\left(2.00\mathrm{kg}\right)\left(3.00\mathrm{m}/\mathrm{s}\right)}{2.00\mathrm{kg}+6.00\mathrm{kg}+2.00\mathrm{kg}} \\ =\frac{6.00\mathrm{kg}.\mathrm{m}/\mathrm{s}}{10.0\mathrm{kg}} \\ =0.60\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of block C and R together is $0.60\mathrm{m}/\mathrm{s}$.

At t = 0.80 s, momentum conservation gives,

$$\begin{gathered} M_c{v}_2+m_R\left({\mathrm{v}}_2+3\mathrm{m}/\mathrm{s}\right)+\left(M_c+m_R\right)v_1 \\ {M}_c{v}_2+m_R{v}_1\left(3m/s\right)=\left(M_c+m_R\right)v_1 \\ \left(m_c+m_R\right)v_2=\left(M_c+m_R\right)v_1-m_R\left(3\mathrm{m}/\mathrm{s}\right) \\ {v}_1=\frac{\left(M_c+m_R\right)v_1-m_R\left(3\mathrm{m}/\mathrm{s}\right)}{\left(M_c+m_R\right)} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{v}}_2=\frac{\left(6.00\mathrm{kg}+2.00\mathrm{kg}\right)\left(0.60\mathrm{m}/\mathrm{s}\right)-\left(2.00\mathrm{kg}\right)\left(3\mathrm{m}/\mathrm{s}\right)}{\left(6.00\mathrm{kg}+2.00\mathrm{kg}\right)} \\ =\frac{\left(4.8\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)-\left(6\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}{\left(8.00\mathrm{kg}\right)} \\ =\frac{1.2\mathrm{kg}.\mathrm{m}/\mathrm{s}}{8.00\mathrm{kg}} \\ =0.15\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocity of block C after second explosion is,${\mathrm{v}}_2=-\left(0.15\right)\hat{i}\mathrm{m}/\mathrm{s}$.

Between t = 0 and t = 0.80 s, the block moves,

$$\begin{gathered} {\mathrm{v}}_1∆\mathrm{t}=\left(0.60\mathrm{m}/\mathrm{s}\right)\times \left(0.80\mathrm{s}-0\right) \\ =0.48\mathrm{m} \end{gathered}$$

Similarly, between t = 0.80 s and t = 2.80 s, it moves an additional,

$$\begin{gathered} {\mathrm{v}}_2∆\mathrm{t}=\left(-0.15\mathrm{m}/\mathrm{s}\right)\times \left(2.80\mathrm{s}-0.80\mathrm{s}\right) \\ =\left(-0.15\mathrm{m}/\mathrm{s}\right)\times \left(2.0\mathrm{s}\right) \\ =-0.30\mathrm{m} \end{gathered}$$

Hence, its net displacement, since, t = 0 is,

$$\begin{gathered} v_1∆t+v_2∆t=\left(0.48-0.30\right)\mathrm{m} \\ =0.18\mathrm{m} \end{gathered}$$

Hence, the position of its center is 0.18 m.

Therefore, by using law of conservation of momentum, the position and velocity of the given object can be found.