Question

A ball of mass 100 g strikes a wall normally with a
speed of 10 m/s and rebounds with the same speed
but in opposite direction. The magnitude of change
in linear momentum of the ball will be
​

Answer 1

Answer:

The change in linear momentum is 2 kg/ms.

Explanation:

Mass of the ball = 100 g

Strikes the wall at speed of = 10 m/s

The ball reverts back with the speed of = -10 m/s (as it is in opposite direction)

★ Change in momentum = Final momentum – Initial momentum

Initial momentum =

• m = 100 g = 0.1 kg
• u = 10 m/s
• v = -10 m/s

$\longrightarrow$ p = m × u

$\longrightarrow$ p = 0.1 × 10

$\longrightarrow$ p = 1 kg/ms

Final momentum =

$\longrightarrow$ p = m × v

$\longrightarrow$ p = 0.1 × -10

$\longrightarrow$ p = -1 kg/ms

★ Change in momentum =

$\longrightarrow$ Final momentum - Initial momentum

$\longrightarrow$ -1 - 1

$\longrightarrow$ -2 kg/ms

Change in momentum is of -2 kg/ms. Negative sign states that the change in momentum is in opposite direction of initial momentum.

Therefore, the change in linear momentum is 2 kg/ms.

Answer 2

Given :-

A ball of mass 100 g strikes a wall normally with a  speed of 10 m/s and rebounds with the same speed  but in opposite direction

To Find :-

Magnitude

Solution :-

We know that

1 kg = 1000 g

100 g = 100/1000 = 0.1 kg

Now

Initial momentum = Mass × Initial velocity

p₁ = m × u

p₁ = 0.1 × 10

p₁ = 1 kg m/s

Final momentum = Mass × Final velocity

Since, It move in opposite direction.

So v = -10 m/s

p₂ = m × v

p₂ = 0.1 × (-10)

p₂ = -1 kg m/s

Now

Δp = p₂ - p₁

Δp = -1 - 1

Δp = -2 kg/ms

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