Question

10. A ball of mass 0.745 kg is dropped from rest from a height of 3.75 m. It rebounds from the ground and reaches a height of 1.78 m, what impulse (specify magnitude and direction) was delivered to the ball by the ground?

Answer 1

10.77kgm/s

Step-by-step explanation:

Impulse is defined as change in momentum of a body.

From Newton's second law

F = ma

F = m(v-u)/t

Ft = m(v-u) = Impulse

where;

M is the mass of the object

v is the final velocity of the object

u is the initial velocity of the object.

Using the formula to time the impulse

Impulse = m(v-u)

m = 0.745kg

Since the ball dropped from rest at a height of 3.75m, it's initial velocity can be calculated according to the equation of motion

g is positive (since it dropped from a height I.e downward motion)

v²= u²+2gH

v² = 0²+2(9.8)(3.75)

v² = 73.5

v = √73.5

v = 8.57m/s(initial velocity of the ball)

If the ball rebounds to a height if 1.78m, g will be negative (since it is an upward motion)

In this case

v² = u²-2gH

0² = u²-2(9.8)(1.78)

-u² = -34.89

u = √34.89

u = 5.91m/s (final velocity of the ball)

Impulse = 0.745(5.91-(-8.57))

Impulse = 0.745(5.91+8.57)

Impulse = 0.745(14.48)

Impulse = 10.77kgm/s

Note that the initial velocity was negated while calculating impulse due to rebounce.