Question

a 0.2 - kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. the balls speed just before and just after impact with the floor is 10 m/s. determine the magnitude of the impulse delivered to the floor by the steel ball.
a. zero n·s
b. 2 n·s
c. 4 n·s
d. 20 n·s
e. 200 n·s

Explanation:

Step1: Define the formula for impulse

Impulse \(J = \Delta p=m\Delta v\)

Step2: Determine the change in velocity

Let the downward direction be negative. Initial velocity \(v_{i}=- 10\ m/s\), final velocity \(v_{f}=10\ m/s\). Then \(\Delta v=v_{f}-v_{i}=10 - (-10)=20\ m/s\)

Step3: Calculate the impulse

Given \(m = 0.2\ kg\), using \(J=m\Delta v\), we have \(J=0.2\times20 = 4\ N\cdot s\)

Answer:

C. \(4\ N\cdot s\)