A baseball is pitched horizontally toward home plate with a velocity of 110 km/h. In which of the following scenarios does the baseball have the largest change in momentum?

(a) The catcher catches the ball.

(b) The ball is popped straight up at a speed of 110 km/h.

(c) The baseball is hit straight back to the pitcher at a speed of 110 km/h.

(d) Scenarios (a) and (b) have the same change in momentum.

(e) Scenarios (a), (b), and (c) have the same change in momentum.

The change in momentum of the baseball can be obtained by the difference between the initial and final momentum.

The baseball's momentum is the product of the mass of the baseball and its velocity.This change in momentum of the baseball will help to evaluate the force acting on the baseball.

When the catcher catches the baseball, the momentum of the baseball becomes zero. So, the change in momentum will be equal to the initial momentum by which the ball is projected toward the catcher initially.

It is given that the speed of the baseball during pop up is\({v_{\rm{x}}} = {v_{\rm{y}}} = 110{\rm{ km/h}}\).

The x-component of the momentum can be calculated as:

\({p_{\rm{x}}} = m{v_{\rm{x}}}\)

The y-component of the momentum can be calculated as:

\({p_{\rm{y}}} = m{v_{\rm{y}}}\)

The resultant momentum when the ball is popped straight up is:

\(\begin{aligned}{c}p = \sqrt {p_{\rm{x}}^{\rm{2}} + p_{\rm{y}}^{\rm{2}}} \\ = \sqrt {{{\left( {m{v_{\rm{x}}}} \right)}^2} + {{\left( {m{v_{\rm{y}}}} \right)}^2}} \\ = m\sqrt {v_{\rm{x}}^{\rm{2}} + v_{\rm{y}}^{\rm{2}}} \end{aligned}\)

Substitute the value\({v_{\rm{x}}} = {v_{\rm{y}}}\)in the above equation.

\(\begin{aligned}{c}p = m\sqrt {v_{\rm{x}}^{\rm{2}} + v_{\rm{x}}^{\rm{2}}} \\p = \sqrt 2 m\sqrt {v_{\rm{x}}^{\rm{2}}} \\ = \sqrt 2 m{v_{\rm{x}}}\end{aligned}\)

This can be expressed as:

\(\begin{aligned}{c}p = \sqrt 2 {p_{\rm{x}}}\\ = \sqrt 2 {p_{\rm{y}}}\end{aligned}\)

The change in momentum is equal to the resultant momentum, whose value is the product of the square root of two and the initial momentum.

The baseball rebounds in the opposite direction, having the same magnitude of momentum as initial momentum when it hits the pitcher.

The change in momentum can be calculated as:

\(p = m\left( {{v_{\rm{f}}} - {v_{\rm{i}}}} \right)\)

Here,\({v_{\rm{f}}}\)is the final velocity of the baseball, which is equal to\(\left( { - {v_{\rm{i}}}} \right)\),\({v_{\rm{i}}}\)is the initial velocity of the baseball, and m is the mass of the baseball.

Substitute the values in the above equation.

\(\begin{aligned}{c}p = m\left( { - {v_{\rm{i}}} - {v_{\rm{i}}}} \right)\\ = - 2m{v_{\rm{i}}}\end{aligned}\)

Here, negative sign indicates the direction for change in momentum.

An impulse force exerted by the baseball is twice the initial momentum of the baseball when it hits the pitcher. So, the magnitude of change in momentum is the largest when the baseball hits straight back to the pitcher.

Thus, option (c) is correct.