A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?

24

Explanation:

The mass = 3 kg

at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.

PE = mgh

but ME = PE + KE = constant   (law of energy conservation)

KE is the kinetic energy

since KE is zero at this point, then,

ME = mgh

where m is the mass

g is acceleration due to gravity = 9.81 m/s^2

h is the height = O

ME = 3 x 9.81 x O

ME = 29.43-O

At point A the total ME is due to its PE and its kE

PE at this point = mgh = 3 x 9.81 x A = 29.43-A

KE =

velocity = v at this point, therefore,

KE = =

therefore,

ME = 29.43-A +

Equating ME for the points O and A, we have

29.43-O = 29.43-A +

29.43-O - 29.43-A =

(O - A)29.43 =

O - A = 0.051   this is the distance between point O and A

For point B

PE = 29.43-B

KE = = 37.5        (velocity is equal to at this point)

therefore,

ME = 29.43-B + 37.5

Equating the ME for points A and B, we have

29.43-A + = 29.43-B + 37.5

29.43-A - 29.43-B = 37.5 -

(A - B)29.43 = 36

A - B = 1.22    this is the distance between points A and B

The distance between points A & B divided by the distance between points 0 & A will be

1.22/0.051 = 23.9 ≅ 24

terd muffin is the correct answer

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s