Respuesta :
Answer:
The speed of the bullet is = 27.36 [tex]\frac{m}{s}[/tex]
Explanation:
Mass of the block (M) = 1 kg
Spring constant = 2500 N per meter
Mass of the bullet ([tex]m_{b}[/tex]) = 20 gm = 0.2 kg
Amplitude A = 0.1 m
Now the final velocity of block when bullet is entered into the block is given by
[tex]V_f = \sqrt{\frac{k}{M + m_b} } A[/tex]
[tex]V_f = \sqrt{\frac{2500}{1 + 0.2} } 0.1[/tex]
[tex]V_f =[/tex] 4.56 [tex]\frac{m}{sec}[/tex]
From conservation of momentum principal
[tex]P_{after} = P_ {before}[/tex]
[tex](M+m_{b} )V_f = m_{b} v_b + M v_B[/tex]
Since [tex]v_B = 0[/tex]
1.2 × 4.56 = 0.2 [tex]v_b[/tex]
[tex]v_b[/tex] = 27.36 [tex]\frac{m}{s}[/tex]
Therefore the speed of the bullet is = 27.36 [tex]\frac{m}{s}[/tex]
The speed of the bullet before the collision is 252.45 m/s.
The given parameters;
- mass of the block, m = 1.0 kg
- spring constant, k = 2500 N/m
- mass of the bullet, m = 20 g = 0.02 kg
- amplitude of the oscillation, A = 10 cm = 0.1 m
The final velocity of the bullet-block system is calculated by applying the principle of conservation of energy;
[tex]K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\mv^2 = kA^2\\\\v^2 = \frac{kA^2}{m} \\\\v = \sqrt{\frac{kA^2}{m} } \\\\v = \sqrt{\frac{(2500) \times (0.1)^2}{(1 + 0.02)} } \\\\v = 4.95 \ m/s[/tex]
The initial velocity of the bullet is calculated by applying the principle of conservation of linear momentum for inelastic collision as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1 + m_2)\\\\[/tex]
where;
- [tex]u_1[/tex] is the initial velocity of the block = 0
- [tex]u_2[/tex] is the initial velocity of the bullet
[tex]1(0) \ + \ 0.02(u_2) = 4.95(1 + 0.02)\\\\0.02u_2 = 5.049\\\\u_2 = \frac{5.049}{0.02} \\\\u_2 = 252.45 \ m/s[/tex]
Thus, the speed of the bullet before the collision is 252.45 m/s.
Learn more about inelastic collision here: https://brainly.com/question/7694106
