A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .
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Answer:
The maximum height reached by the two blocks is approximately 0.1147959 × v₀²
Explanation:
The mass of block B = m
The mass of block A = 3·m
The initial velocity of block B, v₂ = 0 m/s
The initial velocity of block A, v₁ = v₀
The amount of friction between the blocks and the surface = Negligible friction
By the Law of conservation of linear momentum, we have;
Total initial momentum = Total final momentum
3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃
Plugging in the values for the velocities gives;
3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃
∴ 3·m × v₀ = 4·m·v₃
The kinetic energy, K.E. of the combined blocks after the collision is given as follows;
K.E. = 1/2 × mass × v²
The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards
The potential energy, P.E. = m·g·h
Where;
m = The mass of the object at the given height
g = The acceleration due to gravity
h = The height at which the object of mass, 'm', is located
Therefore, for h = The maximum height reached by the two blocks, we have;
P.E. = K.E.
The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².