Question

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 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.

Answer 1

4 N s

Step-by-step explanation:

mass, m = 0.2 kg

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

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

Impulse is defined a the change in momentum .

Impulse = m ( v - u)

Impulse = 0.2 ( 10 + 10)

Impulse = 4 N s

thus, the impulse is 4 N s .

Answer 2

Step-by-step explanation:

Given

mass of steel ball
`m=0.2\ kg`

initial speed of ball
`u=10\ m/s`

Final speed of ball
`v=-10\ m/s` (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

`J=\Delta P`

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

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

`\Delta =-4\ N-s`

so magnitude of Impulse =4 N-s