A 49.5 gram super ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 19.5 m/s. a high-speed camera records this
event. if the ball is in contact with the wall for 4.25 ms, what is the magnitude of the average acceleration of the ball during this time interval
1 answer:
T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)
The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s
The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.
Answer: The magnitude of the acceleration is 11.25 m/s²
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