Respuesta :
Answer:
A) 2.75 m/s B) 0.1911 m C) 0.109 s
Explanation:
mass of block = M =0.1 kg
spring constant = k = 21 N/m
amplitude = A = 0.19 m
mass of bullet = m = 1.45 g = 0.00145 kg
velocity of bullet = vᵇ = 68 m/s
as we know:
Angular frequency of S.H.M = ω₀ = [tex]\sqrt\frac{k}{M}[/tex]
= [tex]\sqrt\frac{21}{0.1}[/tex]
= 14.49 rad/sec
A) Speed of the block immediately before the collision:
displacement of Simple Harmonic Motion is given as:
[tex]x = A sin (\omega t + \phi)\\[/tex]
Differentiating this to find speed of the block immediately before the collision:
[tex]v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\[/tex]
As bullet strikes at equilibrium position so,
φ = 0
t= 2nπ
⇒ cos (ω₀t + φ) = 1
⇒ [tex]v= A\omega_{o}[/tex]
[tex]v=(.19)(14.49)\\v= 2.75 ms^{-1}[/tex]
B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:
S.H.M after collision is given as :
[tex]x= Bsin(\omega t + \phi)[/tex]
To find B, consider law of conservation of energy
[tex]K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}[/tex]
[tex]\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m[/tex]
C) Time taken by the block to reach maximum amplitude after the collision:
Time period S.H.M is given as:
[tex]T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}[/tex]
Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period
[tex]T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec[/tex]
