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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

5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							The correct option is (D) 28 kg. m/s.
Given Data:

Mass of the ball, m = 2 kg
When the ball strikes the floor, the velocity of the ball, v1= 8 m/s
When the ball rebounds upward, the velocity of the ball, v2= -6 m/s
Momentum 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




5 years ago
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