XERCISE
1 answer:
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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 =
=
= 14.49 rad/sec
<h3>A) Speed of the block immediately before the collision:</h3>displacement of Simple Harmonic Motion is given as:
Differentiating this to find speed of the block immediately before the collision:
As bullet strikes at equilibrium position so,
φ = 0
t= 2nπ
⇒ cos (ω₀t + φ) = 1
⇒
S.H.M after collision is given as :
To find B, consider law of conservation of energy
Time period S.H.M is given as:
Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period