Question

If you drop a tennis ball from a height of 2.1 m and it will bounce back after hitting the floor. a) Estimate the mass of the ball and the height to which it bounces back after hitting the floor. [1 mark] b) What was the total impulse on the ball when it hit the floor

Answer 1

Using the law of conservation of mechanical energy and momentum, we can obtain the following results

a) two case:

• case of elastic shock the initial and final height is the same
• case of quasi-elastic shock  the final height is a fraction of the initial height  h_f= e h₀

b) With the relationship between impulse adn momentum variation

The impulse for each case are:

• elastic shock           I = m 12.832

• quasi-elastic shock I = m 6.416 (e + 1)

This is an exercise where we must use the concepts of conservation of mechanical energy and momentum

giving parameter

• initial height  h₀ = 2.1 m

Let's solve the exercise in parts using the conservation of mechanical energy to find the velocity when reaching the ground and the conservation of the momentum for the impulse

a) First part. Let's calculate the speed of the ball when it hits the ground, let's use the conservation of mechanical energy

starting point. Before releasing the ball

        Em₀ = U = m g h

final point. Right when you hit the floor

       Em_f = K = ½ m v²

mechanical energy is conserved

       Em₀ = Em_f

       m g h = ½ m v²

       $v = \sqrt{2gh}$

let's calculate

        v = $\sqrt{2 \ 9.8 \ 2.1}$

        v = 6.416 m/s

Second part. When the ball reaches the ground we have a collision, we can have two cases of elastic collision and quis-elastic collision.

In the case of the elastic collision, the kinetic energy is conserved and the speed of the ball before and after the collision is the same, but in the opposite direction, therefore the initial and final momentum is the same.

In this case the initial mechanistic energy that is kinetic is converted into potential energy, therefore the height at which the ball reaches the same initial height.

           h_f = h₀

This process occurs for any masses of the ball

Second case. quasi-elastic shock

in this case only part of the kinetic energy is conserved, therefore the conservation of momentum is

       $e= \frac{p_f}{p_o}$

where p₀ and p_f are the initial and final momentum, respectively, and e is the coefficient of restitution, which has values ​​between 0 and 1

      $e= \frac{m \ v_f}{m \ v_o}$

      v_f = e v₀

we substitute

      v_f = e 6.416

for this case the conservation of mechanical energy

starting point. Coming off the ground

      Em₀ = K = ½ m v_f²

final point. In the highest part of the climb

      Em_f = U = m g h_f

energy is conserved

       ½ m (e 6.416)² = m g h_f

       $h_f = \frac{1}{2} \frac{e^2 \ 6.416^2}{g}$        

      h_f = e² 2.1

we can see that this height is less than the initial height where the ball was released.

b) momentum is requested in the crash on the floor, the impulse is

         I = Δp

where I is the momentum and Δp the variation of the momentum

solve the two cases

• elastic collision

        I = p_f -p₀

         I = m (v_f - v₀o)

         v_f = -vo

if we choose the vertical direction upwards as positive the initial velocity is negative since it points downwards

         I = -2 m v₀

         I = -m 2 (-6.416)

         I = m 12.832

the sign of the impulse indicates that it is directed upwards

• crash quasi-elastic case

        I = p_f -p₀

        I = m (-e v₀ - v₀)

        I = -m vo (e + 1)

        I = -m 6.46 (e + 1)

You should note that with the data of the exercise the mass must be a data since it cannot be calculated.

Learn more about conservation principles here:

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