Respuesta :
Answer:
(a) The speed of the lighter block is [tex]v_{2x} = 3.7~m/s[/tex].
The speed of the heavier block is [tex]v_{3x} = 3.55~m/s[/tex].
(b) The smaller block goes up to 0.69 m.
Explanation:
We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.
In order to solve this question, I will assume that the smaller mass is initially at rest.
Part A:
We will use conservation of energy.
[tex]K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s[/tex]
This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.
Part B:
Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.
[tex]P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}[/tex]
During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.
[tex]K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2[/tex]
The following relation will be used when combining the two equations:
[tex]v_{2x} = v_2\cos{30^\circ}[/tex]
The following equation is useful for combining the two equations:
[tex]v_{3x} = \frac{2m}{(m+M)}v_{1x} = \frac{2(2.2)}{(2.2 + 6.8)}(8.4\cos(30^\circ)) = 3.55~m/s[/tex]
Therefore from the first equation,
[tex]16 = 2.2v_{2x} + 6.8v_{3x} = 2.2v_{2x} + 6.8(3.55) \\v_{2x} = -3.7~m/s[/tex]
Part C:
We will again use the conservation of energy to find the highest point that the mass can go:
[tex]K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv_{2x}^2 + 0 = 0 + mgH\\\frac{1}{2}(2.2)(-3.7)^2 = (2.2)(9.8)H\\H = 0.69 ~m[/tex]
