Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
Weight = Mass * gravity
= 1470* 9.8 = 14406 N ≈ 14,400 N
According to the Law of conservation of momentum, the total momentum of the system before and after the collision remains same. The momentum may transfer from one object to another as a result of the impact, but the overall momentum of the system remains same.
So, the correct answer is option a.
The total initial momentum of all objects interacting with one another usually equals the total final momentum.
