Answer:
The value is [tex]u_1 = 236 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of bullet is [tex]m_b = 0.0125 \ kg[/tex]
The mass of the block is [tex]M_B = 0.240 \ kg[/tex]
The spring constant is [tex]k = 2.25*10^{3} \ N/m[/tex]
The amplitude is [tex]A= 12.4 \ cm = 0.124 \ m[/tex]
Generally according to the conservation of momentum is
[tex]m_b u_1 + M_B u_2 = (m_b + M_B) v[/tex]
given that the block was at rest we have that
[tex]m_b u_1 = (m_b + M_B) v[/tex]
Now the angular velocity of the both bodies is mathematically represented as
[tex]w = \sqrt{\frac{k}{M_B + m_b} }[/tex]
[tex]w = \sqrt{\frac{ 2.25*10^{3}}{ 0.0125 + 0.240 } }[/tex]
[tex]w = 94.4 \ rad/s[/tex]
Given that the system after collision set into oscillation
The maximum linear velocity of the system after impact is mathematically represented as
[tex]v = A * w[/tex]
[tex]v = 94.4 *0.124[/tex]
[tex]v = 11.7 \ m/s[/tex]
From above equation
[tex]u_1 = \frac{(m_b + M_B ) * v}{m_b}[/tex]
=> [tex]u_1 = \frac{(0.240 + 0.0125) * 11.7}{ 0.0125}[/tex]
=> [tex]u_1 = 236 \ m/s[/tex]
