Respuesta :
The maximum compression in the spring is 0.06883 meters.
Given to us,
mass of bullet, [tex]m= 16.2\ grams[/tex]
velocity of bullet, [tex]v= 850\ meter/second[/tex]
mass of the block, [tex]m_2= 40\ kilograms[/tex]
velocity of block, [tex]v_2= 0\ meter/second[/tex]
spring constant, [tex]k = 1000\ Newton/meter[/tex]
Combined mass after collision,
[tex]\begin{aligned}M&= m_1+m_2\\&= 40.0162\ Kilograms\end{aligned}[/tex]
Using the momentum conservation equation, to find out final velocity of the system(bullet and block combined)
[tex]m_1v_1+m_2v_2= MV\\[/tex]
[tex]0.0162\times 850+40\times 0 = 40.0162\times V\\V= 0.3441 meter/sec[/tex]
Now using the Conservation of Energy (Kinetic energy of the system will be equal to potential energy in the spring), to find out the displacement of the spring [tex](x)[/tex]
[tex]\frac{1}{2}kx^2 = \frac{1}{2}MV^2\\\\\frac{1}{2}\times 1000\times x^2= \frac{1}{2} \times40.0162\times0.3441^2\\\\x^2= 0.004738\\\\x= 0.06883\ meters[/tex]
Hence, the maximum compression in the spring is 0.06883 meters.
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