A 0.24 kg blob of clay is thrown at a wall with an initial horizontal velocity of $16 \mathrm {~m} / \mathrm { s }$ If the clay comes to a stop in $91 \mathrm {~ms} \text {, }$ what is the average horizontal force on the clay due to the wall?

Question

Quizplus · Accepted Answer

The average force on the clay is equal to the change in momentum divided by the time interval. The change in momentum is equal to the final momentum minus the initial momentum. The final momentum is zero, and the initial momentum is equal to the mass of the clay times its initial velocity. Therefore, the change in momentum is equal to the mass of the clay times its initial velocity. The time interval is equal to the time it takes for the clay to come to a stop. Therefore, the average force on the clay is equal to the mass of the clay times its initial velocity divided by the time it takes for the clay to come to a stop. Plugging in the given values, we get:$$F = \frac{mv}{t} = \frac{(0.24 	ext{ kg})(16 	ext{ m/s})}{0.091 	ext{ s}} = 42.7 	ext{ N}$$Therefore, the average horizontal force on the clay due to the wall is 42.7 N, which is closest to choice A, 42 N.

A 024 Kg Blob of Clay Is Thrown at a Wall