V = p / m
V = 1000 / 2.5
V = 400 m/s
hope this helps!
Achieve a full outer shell
The speed of the mass v = 0.884 m/s.
<u>Explanation</u>:
Let
K1 represents the kinetic energy of the mass when it is released,
U1 represents the potential energy of the spring when the mass is released,
K2 represents the kinetic energy of the mass when the spring returns to relaxed length,
U2 represents the potential energy of the spring when the spring returns to relaxed length
The spring is stretched by 0.27 - 0.12 = 0.15 m
K1 = 0
U1 = (1/2)
0.8
(0.15)^2
= 0.009 J
U2 = 0
By conservation of energy,
K2 + U2 = K1 + U1
K2 + 0 = 0 + 0.009 J
K2 = 0.009 J
Let v = speed of the mass
K2 = 1/2
m
v^2
m = 23 g = 0.023 kg
0.009 = 1/2
0.023
v^2
0.009 = 0.0115
v^2
v = √(0.009 / 0.0115)
v = 0.884 m/s.
the question is missing another parameter. we need to know the time taken for the ball to come down. You can use the below formula to find the initial vertical speed.
S = 1/2at2 + ut
distance = 1/2 × acceleration × time2 + intial speed × time