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A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally at 12.0 m/s. The missile embeds itself in the ornament during the collision. What is the tension in the wire immediately after the collision?

Answer :

Explanation:

The given data is as follows.

  Mass of the ornament ([tex]m_{1}[/tex]) = 0.9 kg

  Length of the wire (l) = 1.5 m

 Mass of missile ([tex]m_{2}[/tex]) = 0.4 kg

 Initial speed of missile ([tex]u_{2}[/tex]) = 12 m/s

         r = 1.5 m

According to the law of conservation of momentum,

                   [tex]p_{i} = p_{f}[/tex]  

     [tex]m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v[/tex]

Putting the given values into the above formula as follows.

          [tex]m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v[/tex]

         [tex]0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v[/tex]

              0 + 4.8 = 1.3v

                  v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

            [tex]F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}[/tex]

                       = [tex](0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}[/tex]

                       = 11.80 N

Hence, tension in the wire is calculated as follows.

              T = [tex]F_{c} + (m_{1} + m_{2})g[/tex]

                 = [tex]11.80 N + (0.9 + 0.4) \times 9.8[/tex]

                 = 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

The tension in the wire after collision is 24.54 N.

The given parameters;

  • mass of the ornament = 0.9 kg
  • radius of the wire, r = 1.5 m
  • mass of the missile = 0.4 kg
  • initial speed of the missile = 12 m/s

Apply principle of conservation of linear momentum to determine the final velocity of the system after collision;

[tex]m_1 u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.9(0) + 0.4(12) = v(0.9 + 0.4)\\\\4.8= 1.3v\\\\v = \frac{4.8}{1.3} \\\\v = 3.69 \ m/s[/tex]

The tension in the wire after collision is calculated as follows;

[tex]T = F_c + W\\\\T = ma_c + mg\\\\T = m(a_c + g)\\\\T = (m_1 + m_2)(\frac{v^2}{r} + g)\\\\T = (0.9 + 0.4)(\frac{3.69^2}{1.5} + 9.8)\\\\T = (1.3)(9.08 +9.8)\\\\T = 24.54 \ N[/tex]

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