Question

The force F exerted on a ball during a collision with a wall is given as a function of time t by the equation F(t) = αt - βt2, where α = 400 N/s and β = 4000 N/s2. The ball first contacts the wall at t = 0, and the collision lasts for 0.10 s. What is the magnitude of the change in momentum of the ball?

Group of answer choices


3.33 kg m/s, .67 kg*m/s, 0, 3.6

Answer 1

The magnitude of the change in the momentum of the ball during the collision is 0 kgm/s.

The given parameters:

• Applied force, F(t) = αt - βt²
• α = 400 N/s
• β = 4000 N/s²
• Duration of the collision, t = 0.1 s

The magnitude of the change in the momentum of the ball during the collision is calculated as follows;

change in momentum = impulse  

ΔP = Ft

ΔP = (αt - βt²)t

ΔP = αt² - βt³

ΔP = 400t²  - 4000t³

$\Delta P = 400(0.1)^2 - 4000(0.1)^3\\\\\Delta P = 0 \ kgm/s$

Thus, the magnitude of the change in the momentum of the ball during the collision is 0 kgm/s.

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