Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433