A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts for 1.85 ms, which results in a velocity of 37.0 m/s in the opposite direction. What was the magnitude of the horizontal force?
A) 4.99 kN
B) 0.613 kN
C) 2.80 kN
D) 2.19 kN

Question

A  0.140 kg baseball is thrown horizontally with a velocity of  28.9 m/s. It is struck with a constant horizontal force that lasts for  1.85 ms, which results in a velocity of  37.0 m/s in the opposite direction. What was the magnitude of the horizontal force?
A)  4.99 kN
B)  0.613 kN
C)  2.80 kN
D)  2.19 kN

Quizplus · Accepted Answer

The magnitude of the horizontal force can be calculated using the change in momentum over time, which is the impulse. The change in velocity is from 28.9 m/s to -37.0 m/s (since it's in the opposite direction, we consider it negative), so the total change in velocity is 65.9 m/s. The change in momentum ($\Delta p$) is mass ($m$) times change in velocity ($\Delta v$), so $\Delta p = 0.140 \times 65.9$. The time ($t$) is 1.85 ms, which is $1.85 \times 10^{-3}$ s. Force ($F$) is the change in momentum over time, so $F = \frac{\Delta p}{t}$. Plugging in the numbers, we find the force to be approximately 4.99 kN.

A 0140 Kg Baseball Is Thrown Horizontally with a Velocity of Velocity