- #1

- 68

- 0

## Homework Statement

You are standing on scales which read weight in Newtons. A 0.50 kg ball is dropped from a height of 1 m into your hands. Your hands drop from chest level to waist level during the catch, a distance of about 25 cm. Your mass is 62 kg. Assuming that you decelerate the ball uniformly during the catch, what would be the maximum reading on the scales? (Hint: The scales read 607.6 N before you caught the ball.)

Weight of person = 607.6N

Mass of Ball= 0.50 kg

Displacement of ball when dropped = 1.0 metres

Displacement of ball during deceleration = 0.25 metres

## Homework Equations

Force = Mass * acceleration

Force (weight) = Mass * 9.81

Velocity (final) ^2= Velocity (initial) ^2 + 2 * acceleration * displacement

Vf^2 = Vi^2 + (2 * a * d)

## The Attempt at a Solution

Calculate final velocity of ball after it is first dropped before it is caught:

Vf= x

Vi= 0 m/s

a= 9.81 m/s^2

d= 1.0 metres

Vf^2 = Vi^2 + (2 * a * d)

x^2 = 0^2 + (2*9.81*1)

x^2 = 19.62

x = 4.43 m/s

Calculate acceleration (negative) of ball as it is being caught:

Vf= 0 m/s

Vi= 4.43 m/s

a= x

d= 0.25 metres

Vf^2 = Vi^2 + (2 * a * d)

0^2 = 4.43^2 + (2 * x * 0.25)

0 = 19.62 + (0.50x)

-0.50x = 19.62

x = 39.24

Therefore, the acceleration of the ball as it is being caught is 39.24 metres per second squared downward.

Now, calculate the force of the ball:

F=ma

F=0.5 x 39.24

F=19.62

Now, calculate the maximum weight reading on the scale (which occurs as the ball is being caught):

Weight of man + weight of ball + force of ball during deceleration

607.6 newtons + (0.5)*(9.81) + 19.62 newtons

which is...

607.6 newtons + 4.905 newtons + 19.62 newtons=632.125 newtons

Thus, the maximum scale reading would be 632.125 newtons.

Am i right?