Question

A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00-kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Answer 1

Explanation:

It is known that motion of objects on a vertical is an example of non-uniform motion.

When an object is at the highest point of the circle then for crossing highest point the centripetal force balances the weight of the object.

Therefore,    $\frac{mv^{2}}{r} = mg$

At the highest point of the circle, the minimum speed is as follows.

                 v = $\sqrt{rg}$

                    = $\sqrt{2.80 \times 9.8}$

                    = 5.23 m/s

As the sphere falls from highest to lowest point of circle as follows. According to the law of conservation of energy,

       $K.E_{lowest} = K.E_{highest} + P.E_{highest}$

Expression for potential energy is as follows.

          $P.E_{highest} = mgh$

where,    h = diameter of the circle (2r)

So,              $P.E_{highest} = mgh$

                  $P.E_{highest} = mg(2r)$

      $\frac{1}{2}mu^{2} = \frac{1}{2}mv^{2} + mg(2r)$

where,    u = velocity at the lowest time

Hence, the above equation will be as follows.

               $u^{2} = v^{2} + 4gr$

               u = $\sqrt{v^{2} + 4gr}$

                  = $\sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}$

                  = 11.71 m/s

The dart is embedded into bullet and the collision is inelastic. From the law of conservation of momentum,

       $m_{2}v = (m_{1} + m_{2})u$

                v = $\frac{(m_{1} + m_{2})u}{m_{2}}$

                   = $\frac{(20 + 5) \times 11.71}{5}$

                   = $\frac{292.75}{5}$

                   = 58.55 m/s

Thus, we can conclude that the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.55 m/s.