AP Physics 1 Question 437: Answer and Explanation
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5. Question below refers to the following scenario:
A 2 kg ball with a velocity of -20 m/s collides with the wall and bounces back with a velocity of 10 m/s.
What is the impulse during the collision?
- A. 10 N·s
- B. 20 N·s
- C. 30 N·s
- D. 60 N·s
Correct Answer: D
The impulse is equal to the change in linear momentum, which can be calculated as follows:
J = Δp = mΔv = m(vf - vi) = 2 kg(10 m/s - (-20 m/s)) = 60 N·s