A 0.24 kg blob of clay is thrown at a wall with an initial horizontal velocity of 16 m/s. If the clay comes to a stop in 91 ms, what is the average horizontal force on the clay due to the wall?
A) 35 N
B) 42 N
C) 51 N
D) 26 N

Question

Quizplus · Accepted Answer

The average force can be calculated using the formula $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval. The change in momentum, $\Delta p$, is given by $m \cdot \Delta v$, where $m$ is the mass of the clay (0.24 kg) and $\Delta v$ is the change in velocity (from 16 m/s to 0 m/s, so $\Delta v = -16$ m/s). Thus, $\Delta p = 0.24 \cdot -16 = -3.84$ kg·m/s. The time interval, $\Delta t$, is 91 ms, which is $0.091$ s. Plugging these values into the formula gives $F = \frac{-3.84}{0.091} \approx -42.2$ N. The negative sign indicates the direction of the force is opposite to the initial motion, but since we're asked for the magnitude, the answer is 42 N.

A 024 Kg Blob of Clay Is Thrown at a Wall