Answer:
1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%
Explanation:
For the first part, which is speed just before the crash, we can use energy conservation
Initial. Highest point
Em₀ = U = mg y
Final. Low point just before the crash
Emf = K = ½ m v²
Em₀ = Emf
m g y = ½ m v²
v = √ 2 g y
Let's calculate
v = √ (2 9.8 0.05)
v = 0.99 m / s
1) the moment before the crash is
p₀ = m v
p₀ = 0.221 0.99
p₀ = 0.219 kg m / s
After the collision, the car's speed is zero, so its moment is zero.
p = 0
2) change of momentum
Δp = p - p₀
Δp = 0- 0.219
Δp = -0.219 kg m / s
3) the reason is
Δp / p = 1
In percentage form it is 100%
(aq)
This means it’s aqueous (dissolved in water)
The "penetration of the bullet" is 5 m
<u>Explanation</u>:
A "bullet" with "kinetic energy" of = 400J
A resistive force stops the bullet = 8.00 x 10 N
Work = change in energy
Work = ∆ Kinetic Energy (equation 1)
Work =
(equation 2)
From equations 1 and 2 we have,
= ∆ Kinetic Energy
Where
,
Kinetic Energy = 400 J
F = 8.00 x 10 N
(8.00 x 10 N) d = 400 J
(80 N) d = 400 J

d = 5 m
The penetration of the bullet is 5 m
Keister since the fence after I sent the van the refractory period prevents double counting the same event where is after I passed a van they prevent sensing the patient stimulus it’s after