Answer:
0.4113772 s
Explanation:
Given the following :
Mass of bullet (m1) = 8g = 0.008kg
Initial horizontal Velocity (u1) = 280m/s
Mass of block (m2) = 0.992kg
Maxumum distance (x) = 15cm = 0.15m
Recall;
Period (T) = 2Ļā(m/k)
According to the law of conservation of momentum : (inelastic Collison)
m1 * u1 = (m1 + m2) * v
Where v is the final Velocity of the colliding bodies
0.008 * 280 = (0.008 + 0.992) * v
2.24 = 1 * v
v = 2.24m/s
K. E = P. E
K. E = 0.5mv^2
P.E = 0.5kx^2
0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2
0.5*1*5.0176 = 0.5*k*0.0225
2.5088 = 0.01125k
k = 2.5088 / 0.01125
k = 223.00444 N/m
Therefore,
Period (T) = 2Ļā(m/k)
T = 2Ļā(0.992+0.008) / 233.0444
T = 2Ļā0.0042910
T = 2Ļ * 0.0655059
T = 0.4113772 s
The period of the simple harmonic motion (SHM) is 0.42 s.
The given parameters;
The final velocity of the system after the impact is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.008(280) \ + \ 0.992(0) = v(0.008 \ + \ 0.992)\\\\2.24 = v(1) \\\\v = 2.24 \ m/s[/tex]
The spring constant is calculated as follows;
[tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\k = \frac{mv^2}{x^2} \\\\k = \frac{(0.008 + 0.992))\times 2.24^2}{(0.15)^2} \\\\k = 223 \ N/m[/tex]
The angular speed of the simple harmonic motion (SHM) is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{223}{(0.008 + 0.992)} }\\\\\omega = 14.933 \ rad/s[/tex]
The period of the oscillation is calculated as follows;
[tex]T = \frac{2\pi }{\omega} \\\\T = \frac{2\pi }{14.933} \\\\T = 0.42 \ s[/tex]
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