Question

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Answer 1

Answer:

The velocity at B is  $v = 12.6 \ m/s$  

The velocity at C is  $v_c =10.80 \ m/s$

Explanation:

From the question we are told that

     The mass of the bead is $m_b = 0.368 \ kg$

     The first height is  $h_1 = 4.20 \ m$

      The second height is  $h_2 = 2.14 \ m$

      The initial speed of the bead is  $u = 1.96 \ m/s$

According to the law of energy conservation

       $KE _ A + PE_A = KE_B + PE_B$

Now $KE _ i$ is the kinetic energy at A and it is mathematically represented as

        $KE_A = \frac{1}{2} mu^2$

         $PE_i$ is the potential energy at A which is mathematically represented as

       $PE_A = mg h_1$

       $KE_f$ is the kinetic energy at B which is mathematically represented as

           $KE_B = \frac{1}{2} mv^2$

Where v is the speed of the bead at B

           $PE_B$ is the potential energy at B which equal to 0 because height is  0 at B

So

      $\frac{1}{2} mu^2 + mg h_1 = \frac{1}{2} mv^2$

Making v the subject

       $v = \sqrt{u ^2 + 2gh_1}$

substituting values

     $v = \sqrt{1.96 ^2 + 2* 9.8 * 4.20}$

      $v = 12.6 \ m/s$  

According to the law of energy conservation

       $KE _ B + PE_B = KE_C + PE_C$

So

      $KE_B = \frac{1}{2} m v^2$

      $PE_B = 0$

$KE_C$ is the kinetic energy at c which is mathematically represented as

         $KE_C = \frac{1}{2} * m v_c^2$

 $PE_C$ is the potential energy at C which is mathematically represented as

        $PE_C = mg h_2$

So

      $\frac{1}{2} m v^2 = \frac{1}{2} * m v_c^2 + mg h_2$

     $v^2 = v_c^2 + 2g h_2$

making $v_c$ the subject

     $v_c = \sqrt{12.6^2 - 2* 9.8 * 2.14}$

      $v_c =10.80 \ m/s$