A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, w
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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
False.
The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.
The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.
The Kinetic energy of an object is given by:
Where m is the mass of the object (kg) and v is the velocity of the object (ms-1)
Substituting in:
Therefore the Kinetic energy of the object is 1.35MJ
Where m is the mass of the object (kg) and v is the velocity of the object (ms-1)
Substituting in:
Therefore the Kinetic energy of the object is 1.35MJ