A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
1 answer:
Answer:
Explanation:
According to the parameters provided,
mass of the clay lump, m₁ = 0.05 kg
initial velocity of the lump, u₁ = 12 m/s
mass of the cart, m₂ = 0.15 kg
initial speed of the cart, u₂ = 0
As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:
The resultant speed is v = 3 m/s.
Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.
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Answer:
The area that remains is 4.201 m²
Explanation:
Provided that
AB=D= 4 m (R=2 m)
The area of the half-circle AB is
A=2π
The area of the smaller half-circle is
a=5π/6 + 2√3/4 m²
a=5π/6 + √3/2 m²
<pThus, the remaining area = A - a= 2π - (5π/6 + √3/2) m²
The remaining area is expressed as 2π - (5π/6 + √3/2) m²
Answer:
The force required has a magnitude of 2601.9 N
Explanation:
m = 450 kg
Static friction coefficient μs = 0.73
Kinetic friction coefficient μk = 0.59
The force necessary to initiate movement of the crate is .
Once the crate begins to move, the frictional force decreases to .
To maintain the motion of the crate at a steady velocity, we must lower the pushing force to .
Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.
<pMagnitude of the force