Question

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 1

Explanation:

The given data is as follows.

  Mass of the ornament ($m_{1}$) = 0.9 kg

  Length of the wire (l) = 1.5 m

 Mass of missile ($m_{2}$) = 0.4 kg

 Initial speed of missile ($u_{2}$) = 12 m/s

         r = 1.5 m

According to the law of conservation of momentum,

                   $p_{i} = p_{f}$  

     $m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

          $m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

         $0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v$

              0 + 4.8 = 1.3v

                  v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

            $F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}$

                       = $(0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}$

                       = 11.80 N

Hence, tension in the wire is calculated as follows.

              T = $F_{c} + (m_{1} + m_{2})g$

                 = $11.80 N + (0.9 + 0.4) \times 9.8$

                 = 24.54 N

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