The ball hits the wall with a velocity of 15 m/s, and then rebounds from the wall at velocity of 6 m/s.

Question

 The

ball hits the wall with a velocity of 15 m/s, and then rebounds from the wall at a velocity of 6 m/s.

The me in contact with the wall is 1.5 ms. what is the acceleration of the ball during contact with the wall?

Answer 1

Initial Velocity v_i​ = 15 m/s (toward the wall)

final velocity of the ball v_f= −6 m/s (away from the wall, taking the opposite direction as negative)

Step 1: Calculate the change in velocity (Δv): v_{f -} v_i​ = −6m/s−15m/s = −21m/s

Step 2: The contact time is given in 1.5 milliseconds. Convert the contact time to seconds:

time =1.5ms =1.5×10⁻³s = 0.0015s

Step 3: Use the formula for acceleration (a)

a = Δv​/Δt

Substitute the values:

a = -21m/s

0.0015s

= -14000m/ s²

The negative sign indicates that the acceleration is in the direction opposite to the initial velocity of the ball (deceleration).

Therefore, the acceleration of the ball during its contact with the wall is: − 14000m/ s²

4o