A 0.2 kg ball strikes the floor vertically at 10 m s⁻¹ and rebounds at 8 m s⁻¹. The magnitude of the impulse imparted by the floor on the ball is:
A3.6 N s
B2.0 N s
C1.6 N s
D0.4 N s
Answer & Solution
Correct answer: A. 3.6 N s
Taking upward as positive, impulse = m(v_up − v_down) = 0.2[8 − (−10)] = 0.2×18 = 3.6 N s.
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