Question

A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.04s. What is the acceleration of the ball?

Answer 1

What is acceleration?

$\bullet$ It is defined as the rate of change of its velocity with time.

$\sf{Acceleration = \frac{Change \: in \: velocity}{Time \: taken \: for \: change}}$

That is:

$\boxed{\sf{a = \frac{(v - u)}{t}}}$

Given:

Initial velocity (u) = 6.0 m/s

Final velocity (v) = - 4.4 m/s

Note: Because direction of ball has become opposite.

Time = 0.04 s

We know that:

$\boxed{\sf{Acceleration \:(a)\:= \frac{(v - u)}{t}}}$

So:

$\implies$ $\sf{a = \frac{( - 4.4 - 6.0)}{0.04}}$

$\implies$ $\sf{a = \frac{( - 10.4)}{0.04}}$

After multiplying: $\boxed{\sf{ \frac{ - 10.4}{0.04} \times \frac{100}{100}}}$

Note: In order to make calculations simpler.

Now we get:

$\implies$ $\sf{a = \frac{ - 1040}{4} }$

$\implies$ $\sf{a = - 260 \: {ms}^{2} }$

Therefore:

The acceleration of the ball is $\sf{- 260 \: {ms}^{2} }$

Answer 2

After multiplying:

$\frac{ - 10.4}{0.04} \times \frac{100}{100}$

In order to make calculations simpler.

Now we get:

$a = \frac{ - 1040}{4}$

$a = - 260 \: {ms}^{2}$

Final answer:

$- 260 \: {ms}^{2}$