A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

Question

3 months ago

5

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.

2 answers:

Ostrovityanka [3.2K] · Answer 1 · 2026-03-10T11:09:09+03:00

Answer:

Explanation:

Provided:

mass of the steel ball $m=0.2\ kg$

initial velocity of the ball $u=10\ m/s$

Final velocity of the ball $v=-10\ m/s$ (moving upwards)

The impulse given is determined by the change in the momentum of the object.

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

Thus, the magnitude of the Impulse is 4 N-s.

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kicyunya [3.2K] · Answer 2 · 2026-03-10T11:06:48+03:00

4 0

Answer:

4 N s

Explanation:

mass, m = 0.2 kg

initial velocity, u = -10 m/s (downward direction)

final velocity, v = +10 m/s (upward direction)

Impulse is defined as the change in momentum.

Impulse = m ( v - u)

Impulse = 0.2 (10 + 10)

Impulse = 4 N s

Hence, the impulse equals 4 N s.