Question

A ball of mass 0.05 kg strikes a smooth wall normally four times in 2 second with a velocity of 10m/s. Each time the ball rebounds with the same speeds of 10m/s. Calculate the average force on the wall

Answer 1

To calculate the average force exerted on the wall by the ball, use the change in momentum for a single collision and then account for the number of collisions within the given time frame.

Step-by-step explanation:

The student is asking about the calculation of the average force exerted on a wall by a ball hitting it multiple times and rebounding with the same speed. To find the average force, one must understand the concept of impulse and change in momentum. The force can be calculated using the formula: F = ∆p / ∆t, where F is the average force, ∆p is the change in momentum, and ∆t is the time during which the change occurs.

Since the ball strikes the wall four times in 2 seconds, we can find the total change in momentum and divide it by the total time to calculate the average force. For each collision, the change in momentum is 2mv (since it goes from v to -v and ∆p = m(v - (-v)) = 2mv). Multiplying this by the number of collisions and dividing by the total time will provide the average force.

Answer 2

Answer" The average force on the wall be equal to 0.5 N.

Explanation:

To calculate the average force on the wall, we use the following equation:

`Ft=m\Delta v\\\\F=(m(v_2-v_1))/(t)`

where,

F = average force = ?N

m = mass of the ball = 0.05kg

`v_1` = Initial velocity of the ball = -10m/s (negative sign because the ball rebounds)

`v_2` = Final velocity of the ball = 10m/s

t = time taken by the ball = 2s

Putting values in above equation, we get:

`F=(0.05kg[10-(-10)]m/s)/(2s)\\\\F=(0.05kg(20m/s))/(2s)\\\\F=0.5kgm/s^2=0.5N`

Hence, the average force on the wall will be equal to 0.5N.