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2 years ago

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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.

2 answers:

nadya68 [22]2 years ago

7 0

Answer:

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

MAVERICK [17]2 years ago

4 0

Answer:

4 N s

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 .

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