A 0.140-kg baseball is thrown with a velocity of $27.1 \mathrm {~m} / \mathrm { s }$ It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact? A) 1.79 × 10^-3 s B) 1.28 × 10^-2 s C) 3.07 × 10^-2 s D) 4.30 × 10^-3 s

Question

Quizplus · Accepted Answer

The change in momentum is calculated using the formula: $ \Delta p = m(v_f - v_i) $. The impulse is equal to the change in momentum and is also equal to the average force times the contact time: $ F \Delta t = \Delta p $. Solving for $ \Delta t $ gives $ \Delta t = \frac{m(v_f - v_i)}{F} = \frac{0.140 \times (37.0 - (-27.1))}{5000} = 1.79 \times 10^{-3} $ s.

A 0140-Kg Baseball Is Thrown with a Velocity Of Of 27.1