Centre of Mass then axis of rotation and then moment of inertia. This was the toughest question for your level... happy to help ^_^. It was purely experimental question.
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know
where:
a= acceleration
we have:
initial velocity,
Using the eq. of motion:
where:
v= final velocity after the separation of spring with the bullet.
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
∵At maximum height the final velocity will be zero
Using the equation of motion:
where:
h= height
g= acceleration due to gravity
is the height from the release position of the spring.
So, the height from the latched position be:
Answer:
F = m a = m v / t where v is the change in velocity in time t
F = p / t since m v is equal to p
F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N
Or you can use the impulse equation
Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
