A 0.140-kg baseball is thrown with a velocity of $27.1 \mathrm {~m} / \mathrm { s }$ It is struck by the bat exerting 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 impulse-momentum theorem states that the change in momentum equals the impulse applied. The change in velocity is $27.1 + 37.0 = 64.1 \, \mathrm{m/s}$. The change in momentum is $0.140 \, \mathrm{kg} \times 64.1 \, \mathrm{m/s} = 8.974 \, \mathrm{kg \cdot m/s}$. The impulse is $5000 \, \mathrm{N} \times t$. Solving $5000t = 8.974$ gives $t = 1.79 \times 10^{-3} \, \mathrm{s}$.

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