Question

A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mass of 200 g. The ball is lifted to a height of 1.5 m above the ground and then released from rest. The ball swings to its lowest point where the string breaks. The ball is then in free-fall until it hits the ground. How far would the ball travel in the horizontal direction between points B and C (i.e. what is the range)?

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is $D = 1.38 m$

Explanation:

From the question we are told that

      The length of the string is  $L = 1.6 \ m$

      The mass of the ball is  $m = 200 g = \frac{200}{1000} = 0.2 \ kg$

       The height of ball is  $h = 1.5 \ m$

Generally the work energy theorem can be mathematically represented as

               $PE = KE$

   Where PE is the loss in potential energy which is mathematically represented as

                   $PE =mgh$

Where h is the difference height of ball at A and at B  which is mathematically represented as

                 $h = y_A - y_B$

So        $PE =mg(y_A - y_B)$              

While KE is the gain in kinetic energy which is mathematically represented as

               $KE = \frac{1}{2 } (v_b ^2 - 0)$

Where $v_b$ is the velocity of the of the ball

  Therefore we have from above that

                    $PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)$

               Making $v_b$ the  subject we have

      $v_b = \sqrt{2g (y_A - y_B)}$

substituting values

      $v_b = \sqrt{2g (1.5 - 0.40)}$

     $v_b = 4.6 \ m/s$

Considering velocity of the ball when it hits the  floor in terms of its vertical and horizontal component we have

         $v_x = 4.6 m/s \ while \ v_y = 0 m/s$

The time taken to travel  vertically from the point the ball broke loose  can be obtained using the equation of motion

            $s = v_y t - \frac{1}{2} g t^2$

Where s is distance traveled vertically which given in the diagram as $s = -0.4$

The negative sign is because it is moving downward

     Substituting values

              $-0.4 = 0 -\frac{1}{2} * 9.8 * t^2$

         solving for t we have  

               $t = 0.3 \ sec$

Now the distance traveled on the horizontal is mathematically evaluated as

           $D = v_b * t$

           $D = 4.6 * 0.3$

           $D = 1.38 m$