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
[tex]\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0[/tex]
The kinetic energy, K.E. of the combined blocks after the collision is given as follows;
K.E. = 1/2 × mass × v²
[tex]\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2[/tex]
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.
[tex]m \cdot g \cdot h = \dfrac{9}{8} \cdot m\cdot v_0^2[/tex]
[tex]h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot v_0^2 \approx 0.1147959 \cdot v_0^2[/tex]
The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².
