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A 2-kg ball moving straight down strikes the floor at 8 m/s. It rebounds upward at 6 m/s, What is the magnitude of the change in momentum of the ball? (A) 2 kg· m/s (B) 4 kg' m/s (C) 14 kg· m/s (D) 28 kg' m/s
The correct option is (D) 28 kg. m/s.Given Data:Mass of the ball, m = 2 kgWhen the ball strikes the floor, the velocity of the ball, v1= 8 m/sWhen the ball rebounds upward, the velocity of the ball, v2= -6 m/sMomentum of a particle (p) is equal to the mass of the particle (m) times velocity of the particle (v).So, p = mv …… (1)When the ball strikes the floor having mass 2 kg (m=2 kg) and velocity 8 m/s (v1=8 m/s), the momentum of the ball (p1) will be,p1= mv1 = (2 kg) (8 m/s) = 16 kg. m/s ……. (2)When the ball rebounds upward having mass 2 kg (m=2 kg) and velocity 6 m/s (v1= -6 m/s), the momentum of the ball (p2) will be,p2= mv2 = (2 kg) (-6 m/s) = -12 kg. m/s …… (3)So the magnitude of the change in momentum will be,Change in momentum = p1- p2 = 16 kg. m/s – (-12 kg. m/s) = 28 kg. m/s …… (4)From equation (4) we observed that, the magnitude of the change in momentum of the ball will be, 28 kg. m/s. Thus option (d) is correct
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