Step 1

We are given that the mass of bullet is $m$, the initial velocity of the bullet is ${v_0}$, the mass of target is $M$ and the stiffness constant of each spring is $k$.

We are asked to obtain the critical damping coefficient and the maximum compression of the spring.

 
Step 2

The springs are attached in the parallel combination, the equivalent stiffness constant of the springs is calculated as:

\[\begin{array}{c} {k_{eq}} = k + k\\ = 2k \end{array}\]
 
Step 3

When the bullet is attached with the target then the total mass of the system is given as:

\[{M_0} = m + M\]
 
Step 4

To calculate the natural frequency of the vibration we use the formula:

\[{\omega _n} = \sqrt {\frac{{{k_{eq}}}}{{{M_0}}}} \]
 
Step 5

Substitute the known value in the formula:

\[{\omega _n} = \sqrt {\frac{{2k}}{{m + M}}} \]
 
Step 6

To calculate the critical damping coefficient of the system we use the formula:

\[{c_c} = 2{M_0}{\omega _n}\]
 
Step 7

Substitute the known values in the formula:

\[\begin{array}{c} {c_c} = 2\left( {m + M} \right)\sqrt {\frac{{2k}}{{m + M}}} \\ = \sqrt {8\left( {m + M} \right)k} \end{array}\]
 
Step 8

To obtain the equation of motion for the critically damped system we use the equation:

\[x = \left( {A + Bt} \right){e^{ - {\omega _n}t}}\]…… (1)
 
Step 9

At $t = 0$, $x = 0$, therefore the above equation becomes:

\[\begin{array}{c} 0 = \left( {A + B\left( 0 \right)} \right){e^{ - {\omega _n}\left( 0 \right)}}\\ = A\left( 1 \right)\\ A = 0 \end{array}\]
 
Step 10

Differentiate the equation (1) with respect to time:

\[\begin{array}{c} \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {A + Bt} \right){e^{ - {\omega _n}t}}\\ v = \left( {0 + B\left( 1 \right)} \right){e^{ - {\omega _n}t}} + \left( {A + Bt} \right){e^{ - {\omega _n}t}}\left( { - {\omega _n}} \right)\\ = \left( B \right){e^{ - {\omega _n}t}} + \left( {0 + Bt} \right){e^{ - {\omega _n}t}}\left( { - {\omega _n}} \right)\\ = \left( {1 - {\omega _n}t} \right)B{e^{ - {\omega _n}t}} \end{array}\]…… (2)
 
Step 11

According to conservation of linear momentum:

\[\begin{array}{c} m{v_0} = \left( {M + m} \right)v\\ v = \frac{{m{v_0}}}{{M + m}} \end{array}\]
 
Step 12

At $t = 0$, $v = \frac{{m{v_0}}}{{M + m}}$, therefore the equation (2) becomes:

\[\begin{array}{c} \frac{{m{v_0}}}{{M + m}} = \left( {1 - {\omega _n}\left( 0 \right)} \right)B{e^{ - {\omega _n}\left( 0 \right)}}\\ = \left( 1 \right)B\left( 1 \right)\\ B = \frac{{m{v_0}}}{{M + m}} \end{array}\]
 
Step 13

Substitute the known values in equation (1):

\[\begin{array}{c} x = \left( {0 + \left( {\frac{{m{v_0}}}{{M + m}}} \right)t} \right){e^{ - {\omega _n}t}}\\ = \left( {\frac{{m{v_0}}}{{M + m}}} \right)t{e^{ - {\omega _n}t}} \end{array}\]…… (3)
 
Step 14

Substitute the known values in equation (2):

\[v = \left( {1 - {\omega _n}t} \right)\left( {\frac{{m{v_0}}}{{M + m}}} \right){e^{ - {\omega _n}t}}\]
 
Step 15

For maximum compression of the system, $v = 0$, therefore, the above equation becomes:

\[\begin{array}{c} 0 = \left( {1 - {\omega _n}t} \right)\left( {\frac{{m{v_0}}}{{M + m}}} \right){e^{ - {\omega _n}t}}\\ {e^{ - {\omega _n}t}}\left( {1 - {\omega _n}t} \right) = 0 \end{array}\]
 
Step 16

The first term of the above equation cannot be zero as time will become infinite for the condition therefore, the second term is equal to zero which is solved as:

\[\begin{array}{c} 1 - {\omega _n}t = 0\\ {\omega _n}t = 1\\ t = \frac{1}{{{\omega _n}}} \end{array}\]
 
Step 17

Substitute the known value in equation (3):

\[\begin{array}{c} x = \left( {\frac{{m{v_0}}}{{M + m}}} \right)\left( {\frac{1}{{{\omega _n}}}} \right){e^{ - \left( {{\omega _n}} \right)\left( {\frac{1}{{{\omega _n}}}} \right)}}\\ = \frac{{m{v_0}}}{{{\omega _n}\left( {M + m} \right)}}{e^{ - 1}}\\ = \frac{{m{v_0}}}{{{\omega _n}e\left( {M + m} \right)}} \end{array}\]
 
Step 18

Substitute the known value in the above equation:

\[\begin{array}{c} x = \frac{{m{v_0}}}{{\left( {\sqrt {\frac{{2k}}{{M + m}}} } \right)e\left( {M + m} \right)}}\\ = \frac{{m{v_0}}}{{e\sqrt {2k\left( {M + m} \right)} }} \end{array}\]