Answer:
The average net force on the truck is 375 Newtons.
Explanation:
Using Newton's 3rd equation of motion, we have :
×a×s
where, v = final velocity = 25 m/s
u = initial velocity = 20 m/s
a = acceleration
s = distance traveled = 300 m
Using these values in the above equation, we get acceleration = 0.375 m/
Using Newton's second law, we have:
F=m×a
where m = mass = 1000 kg
a= acceleration = 0.375 m/
Putting values we have F=375 N
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%