Where are the statements? You forgot to attach them lol
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer: The ball (option A)
Explanation: change in momentum is defined by the formulae m(v - u) where m = mass of object, v = final velocity and u = initial velocity.
For the ball, it hits the ground and bounces back with the same speed, that's final velocity equals initials (v = - u)
Change in momentum = m( -u- u) = m(-2u) = m(-2u) = -2mu
For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)
m(v - u) = m(0 - u) = - mu.
-2mu (change in momentum from the ball) is greater than - mu ( change in momentum of clay)
