If the marble bounces straight upward to a height of 0.64 m, what is the magnitude of the impulse delivered to the marble by the floor?

0.133kg.m/s

Explanation:

Using equation of motion

v² = u² + 2as where v is the final velocity in m/s,  a = g = 9.81 m/s² and s is the height and u = 0

v² = 2×9.81×1.44m

v = √(2×9..81×1.44) = 5.312 m/s  taking downward movement as negative

v = -5.312 m/s

the marble bounces straight upward to a height of 0.64 m

again

v² = u² + 2as  where final velocity = 0

-u²  = (2 × -9.81 × 0.64)

u = √12.5568 = 3.544m/s

impulse = m × Δv = 0.15 × ( 3.544 --5.312 ) = 0.15 × (3.544+5.312) = 0.133kg.m/s

Given,

The initial position of the boat, d_i=(5i-2j+4k) m

The initial velocity of the boat, u=(4i+j-3k) m/s

The time period, t=8 s

The final velocity of the boat, v=(12i-7j+13k) m/s

To find:

a) The position of the object after 8 s

b) The acceleration of the object.

Explanation:

a)

From the equation of motion, the total distance traveled by the object is given by,

On substituting the known values,

Thus the final position of the boat is,

b)

The acceleration of an object is given by,

On substituting the known values,

Final

a) Thus the position of the boat after 8 s is

The acceleration of the boat is