Question

A3 kg ball moving straight down strikes the floor at 8 m/s. It rebounds upwards at 6 m/s the magnitude of change in momentum is
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Answer 1

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.

Answer 2

$\\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• Mass of ball (m) = 3kg

• Initial velocity (u) = 8m/sec

• Final velocity (v) = -6m/sec

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Change in momentum.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: know:} }}}$

$\\ \boxed { \bf \: momentum =mass \times velocity}$

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

Initial momentum (P) = mu

Initial momentum (P) = 3×8

Initial momentum (P) = 24 kg-m/sec

____________________________

Final momentum (P') = mv

Final momentum (P') = 3×6

Final momentum (P') = 18 kg-m/sec

____________________________

♦ Change in momentum

ㅤㅤㅤㅤㅤ= Final momentum - Initial momentum

ㅤㅤㅤㅤㅤ = P' - P

ㅤㅤㅤㅤㅤ = 18 - 24

ㅤㅤㅤㅤㅤ = -6

$\\ \small \therefore \sf \: change \: in \: momentum \: is \: - 6 \: kg \: m {sec}^{ - 1} .$

Negative sign indicates the direction.

Magnitude is 6 kg m/sec.