Question

A 0.15-kg baseball is pitched with a speed of \(35 \mathrm{m} / \mathrm{s}\) \((78 \mathrm{mph}) .\) When the ball hits the catcher's glove, the glove moves back by \(5.0 \mathrm{cm}(2 \text { in. })\) as it stops the ball. (a) What was the change in momentum of the baseball? (b) What impulse was applied to the baseball? (c) Assuming a constant acceleration of the ball, what was the average force applied by the catcher's glove?

Short answer

The change in momentum of the baseball is -5.25 kg⋅m/s. b) What is the impulse applied to the baseball? The impulse applied to the baseball is -5.25 kg⋅m/s. c) What is the average force applied by the catcher's glove? The average force applied by the catcher's glove is -183.75 N.

Step-by-step solution

Convert the distance to meters

The distance the glove moves back is given in centimeters. To work with SI units, we need to convert this distance to meters: $$5.0 \, \mathrm{cm} \times \frac{1\, \mathrm{m}}{100 \, \mathrm{cm}} = 0.05 \, \mathrm{m}$$

Calculate the change in momentum

To find the change in momentum, we need to first find the final velocity of the ball, which is \(0 \, \mathrm{m/s}\) (the ball stops). Let \(v_i = 35 \, \mathrm{m/s}\) and \(v_f = 0 \, \mathrm{m/s}\). Then, we can calculate the change in momentum using the formula \(\Delta p = m \Delta v\): $$\Delta p = m(v_f - v_i) = (0.15\, \mathrm{kg})(0 - 35\, \mathrm{m/s}) = -5.25 \, \mathrm{kg\cdot m/s}$$

Calculate the impulse applied to the baseball

The impulse applied to the baseball is equal to the change in its momentum. Therefore, we have: $$I = \Delta p = -5.25 \, \mathrm{kg\cdot m/s}$$

Find the average force applied by the catcher's glove

To find the average force applied by the catcher's glove, we need to first find the acceleration and the time interval during which the force is applied. We know that the final velocity (\(v_f\)) is 0, and the initial velocity (\(v_i\)) is 35 m/s. We can find the time interval (\(\Delta t\)) using the equation \(v_f = v_i + a \Delta t\) and the fact that the glove moves back by 0.05 m. We can rewrite the equation as: $$\Delta t = \frac{v_f - v_i}{a}$$ Adding another equation about the displacement of the glove, we have: $$0.05\, \mathrm{m} = v_i \Delta t + \frac{1}{2}a \Delta t^2$$ We can now solve these two equations simultaneously for \(\Delta t\) and \(a\). Multiplying the first equation by \(a\), we have: $$a \Delta t = v_i - v_f$$ Substituting this expression for \(a \Delta t\) into the second equation, we get: $$0.05\, \mathrm{m} = v_i \left(\frac{v_i - v_f}{a}\right) + \frac{1}{2}a \left(\frac{v_i - v_f}{a}\right)^2$$ After simplifying the above equation and solving for \(a\), we find: $$a = -1225 \, \mathrm{m/s^2}$$ Finally, we can find the average force applied by the catcher's glove using the equation \(F = ma\): $$F = (0.15\, \mathrm{kg})(-1225\, \mathrm{m/s^2}) = -183.75 \, \mathrm{N}$$ So, the average force applied by the catcher's glove is -183.75 N. The negative sign indicates that the force is acting in the opposite direction of the initial velocity of the ball.