Question

A ball hits a wall horizontally at 6.0 m/s. It rebounds horizontally at 4.4 m/s. The ball is in contact with the wall for 0.040 s. What is tje acceleration of the ball?​

Answer 1

Given:

• Initial velocity of the ball = 6.0 m/s or 6 m/s

• Final velocity of the ball = – 4.4 m/s (As it rebounds)

• Time (t) = 0.040 s

To calculate:

• Acceleration of the ball (a)

Calculation:

Here, as per the given question we are given that the ball hits a wall horizontally at 6.0 m/s i.e, its initial velocity is 6.0 m/s. Then it rebounds, this means that its direction changes from forward to backward. If we are taking forward direction as positive then backward direction will be negative. So, its final velocity is -4.4 m/s. Also, given that the is in contact with the wall for 0.040 s. Now, as we are given the values, so we'll substitute the values in the formula of acceleration to find the acceleration of the ball.

Points to note:

✰ Acceleration is the rate of change in velocity with time.

✰ It is a vector quantity.

✰ Acceleration = (Final velocity - Initial velocity)/ time

✰ It is denoted as "a".

✰ SI unit of acceleration is m/s².

Now, coming to the formula!

→ a = $\sf { \dfrac{v-u}{t}}$

• a = Acceleration

• v = Final velocity

• u = Initial velocity

• t = Time

Substituting values,

→ a = $\sf { \dfrac{-4.4 - (+6) }{0.040}}$ m/s²

→ a = $\sf { \dfrac{-4.4 - 6 }{0.040}}$ m/s²

→ a = $\sf { \dfrac{-10.4 }{0.040}}$ m/s²

→ a = $\sf { \dfrac{-104000 }{400}}$ m/s²

→ a = $\sf { \dfrac{-1040 }{4}}$ m/s²

→ a = – 260 m/s²

Therefore, the acceleration of the ball is – 260 m/s².