Question

A glass marble, whose mass is 1/10 kg falls from a
height of 2.5 m, and rebounds to a height of 1.6m.
If the time during which they are in contact be one
tenth of a second, then the average force between
the marble and floor is
(1) 12.6 N
(2) 17 N
(3) 17.2 N
(4) 18.4 N​

Answer 1

this is the best i can do

Answer 2

Given:

The mass of the marble = 1/10 kg

The falling height h = 2.5m

The rebounding height h' = 1.6m

The time of contact t = 1/10 s

To Find:

The average force between  the marble and floor

Solution:

The average force between  the marble and floor is  (1) 12.6 N.

Let vi and vf be the velocities of the ball before and after striking the floor respectively.

Conserving mechanical energy before the collision,

mgh = mvi² / 2

or vi = $\sqrt{2gh}$

Taking g = 10 m/s²,

vi = √(2 X 10 X 2.5)

= $5\sqrt{2}$ m/s

Conserving mechanical energy after the collision,

mgh' = mvf² / 2

or vf = $\sqrt{2X10X1.6}$

= 4√2 m/s

The contact force = dp / dt

where p is the momentum

or F = dp / dt

Final momentum = m vf =  4√2/10

Initial momentum = -5√2/10

Time inetrval = 1/10

Substituting,

F = $\frac{\frac{4\sqrt{2} -( -5 \sqrt{2} )}{10} }{0.1}$

= 9√2N

Taking √2 = 1.4,

F = 12.6N