Respuesta :
A block oscillates on the end of the spring. The amplitude of SMH after the collision is 40.67 m.
According to the rule of conservation of momentum, the total amount of momentum before and after a collision is equal.
Given that, m₁ = 2 kg
m₂ = 4 kg
Time period T = 20 ms = 0.2 s
v₁ = 6 m/s
t = 5 ms = 0.005 s
x = 1.0 cm cos (ωt + π/2)
Kinetic energy in block 1 is Ek = 1/2* m₁ v₁²
Energy in spring in block 2 Es = 1/2* k x²
We know, f = 1 / T = 1 / 0.02 = 50 hz
ω = 2 π * f = 2 π * 50 = 100 π
To find k, we know the formula, k = ω² * m₂ = (100 π)² * 2 = 197317
x = (1.0 cm) cos (ωt + π/2)
ω = 100 π rad, t = 0.005 s
x = (0.01) cos (100 π* 0.005 + π/2) = -0.01 m
Energy of block 1 = 0.5* 4* 6 = 72 J ---(1)
Energy of the spring = 0.5* 2* (100 π)² - 0.01² = 9.86 J ---(2)
Energy before collision will be (1) + (2) = e₁ + e₁ = 72 + 9.86 = 81.86 J
Energy after the collision E after = 1/2* k x²
1/2 * k x² = 1/2 * ω² * x²
m = m₁ + m₂ = 6 kg
ω = √(k/m²) = (197317/36) = 74.03 rad/s
Applying, Energy before collision = Energy after collision
81.86 = 1/2 * 6 * 74.03 * x²
x² = 0.37
x = 0.61 m
Amplitude can now be found since we have gotten all the parameters
x = A cos (ω t + π / 2)
0.61 = A cos [181.3 * (0.005) + π / 2]
A = - 0.61/0.015 = 40.67 m
To know more about amplitude:
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