Question

9. During an egg toss, a catcher must cushion the egg by maximizing the time it takes to stop the

egg. Find the impulse applied to a 60.0-gram egg, if the egg has an initial velocity of 6.0 m/s as it

first touches the catcher's hand, and the egg is stopped within 50.0 cm.

Answer 1

the impulse experienced by the egg is 0.053 kgm/s.

Step-by-step explanation:

Given;

mass of the egg, m = 60 g = 0.06 kg

initial velocity of the egg, u = 6 m/s

height moved by the egg, h = 50 cm = 0.5 m

Determine the final velocity of the egg as it moves upward;

v² = u² + 2(-g)h

v² = u² - 2gh

where;

v is the final velocity

-g is negative acceleration due gravity as it moves upward

v² = 6² - 2(9.8 x 0.5)

v² = 26.2

v = √26.2

v = 5.12 m/s

The impulse applied to the egg is the change in linear momentum;

J = ΔP

ΔP = mu - mv

ΔP = m(u - v)

ΔP = 0.06(6 - 5.12)

ΔP = 0.053 kgm/s

Therefore, the impulse experienced by the egg is 0.053 kgm/s.