Answer:
The velocity at B is [tex]v = 12.6 \ m/s[/tex]
The velocity at C is [tex]v_c =10.80 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bead is [tex]m_b = 0.368 \ kg[/tex]
The first height is [tex]h_1 = 4.20 \ m[/tex]
The second height is [tex]h_2 = 2.14 \ m[/tex]
The initial speed of the bead is [tex]u = 1.96 \ m/s[/tex]
According to the law of energy conservation
[tex]KE _ A + PE_A = KE_B + PE_B[/tex]
Now [tex]KE _ i[/tex] is the kinetic energy at A and it is mathematically represented as
[tex]KE_A = \frac{1}{2} mu^2[/tex]
[tex]PE_i[/tex] is the potential energy at A which is mathematically represented as
[tex]PE_A = mg h_1[/tex]
[tex]KE_f[/tex] is the kinetic energy at B which is mathematically represented as
[tex]KE_B = \frac{1}{2} mv^2[/tex]
Where v is the speed of the bead at B
[tex]PE_B[/tex] is the potential energy at B which equal to 0 because height is 0 at B
So
[tex]\frac{1}{2} mu^2 + mg h_1 = \frac{1}{2} mv^2[/tex]
Making v the subject
[tex]v = \sqrt{u ^2 + 2gh_1}[/tex]
substituting values
[tex]v = \sqrt{1.96 ^2 + 2* 9.8 * 4.20}[/tex]
[tex]v = 12.6 \ m/s[/tex]
According to the law of energy conservation
[tex]KE _ B + PE_B = KE_C + PE_C[/tex]
So
[tex]KE_B = \frac{1}{2} m v^2[/tex]
[tex]PE_B = 0[/tex]
[tex]KE_C[/tex] is the kinetic energy at c which is mathematically represented as
[tex]KE_C = \frac{1}{2} * m v_c^2[/tex]
[tex]PE_C[/tex] is the potential energy at C which is mathematically represented as
[tex]PE_C = mg h_2[/tex]
So
[tex]\frac{1}{2} m v^2 = \frac{1}{2} * m v_c^2 + mg h_2[/tex]
[tex]v^2 = v_c^2 + 2g h_2[/tex]
making [tex]v_c[/tex] the subject
[tex]v_c = \sqrt{12.6^2 - 2* 9.8 * 2.14}[/tex]
[tex]v_c =10.80 \ m/s[/tex]
