Question

A ball of moist clay falls 15.0m to the ground. It is in contact with the ground for 20ms before stopping?

(a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground?
(b) Is the average acceleration up or down?

Answer 1

(a) `a_"ave" = 857 m/s^2`

(b) The acceleration was upwards.

Explanation:

To find the velocity when it reached the ground, use

`v^2 = u^2 + 2*a*s`

`v^2 = 0^2 + 2*9.8 m/s^2*15.0 m`

`v^2 = 294 m^2/s^2`

`v = 17.15 m/s`

(a) To find the average acceleration over the time of 0.020 s, use

`v = u + a*t`

Since `17.15 m/s` was the initial velocity and it was downward, let the `u = -17.15 m/s`.

`0 = -17.15 m/s + a*0.020 s`

`a_"ave" = 17.15 m/s /0.020 s = 857 m/s^2`

(b) Since `a_"ave"` turned out to be positive and the falling velocity was negative, the acceleration must have been upwards. And that makes sense, since to stop something, you need a force in the direction opposite the existing velocity. And from `vecF=m*veca`, we can see that acceleration is in the same direction as the force acting on the object.

I hope this helps,
Steve

Answer 2

The velocity of the clay ball as it hits the ground is found from the kinematic expression

`v^2 - u^2 = 2gh`

Substituting given values we get

`v^2 - 0^2 = 2xx9.81xx15.0`
`=>v=17.16\ ms^-1`

After it hits ground it is brought to stop. Acceleration can be found from the kinematic expression

`v=u+at`

Substituting various values we get

`0=17.16+a_"ave"xx(20xx10^-3)`
`=>a_"ave"=-17.16/(20xx10^-3)`
`=>a_"ave"=-857.8\ ms^-2`

(a) `857.8\ ms^-2`
(b) `-ve` sign shows us that it is retardation and opposes the motion. The motion of the ball is in the downward direction. Hence, average acceleration acts in up direction.