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: Recall the impulse - momentum theorem

The impulse \(J\) is equal to the change in momentum \(\Delta p\). The formula for momentum is \(p = mv\). Let the downward direction be negative. The initial velocity \(v_{i}=- 10\ m/s\) (before impact) and the final velocity \(v_{f}=10\ m/s\) (after impact). The mass \(m = 0.2\ kg\).

Step2: Calculate the change in momentum

\(\Delta p=m(v_{f}-v_{i})\). Substitute \(m = 0.2\ kg\), \(v_{i}=-10\ m/s\) and \(v_{f}=10\ m/s\) into the formula:

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Answer:

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