A 2.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s. What is the impulse exerted on the ball by the wall?

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oladayoademosu oladayoademosu · Answer 1 · 2021-01-15T07:33:19+01:00

Answer:

-16i kgm/s

Explanation:

Impulse I = m(v - u) where m = mass of ball = 2.0 kg, u = initial velocity of ball = (4i + 3j) m/s and v = final velocity of ball = (-4i + 3j) m/s.

So, the impulse is thus

I = m(v - u)

= 2.0 kg[(-4i + 3j) m/s - (4i + 3j) m/s]

= 2.0 kg[(-4i - 4i) + (3j - 3j) m/s]

= 2.0kg[-8i + 0j] m/s

= -16i kgm/s

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asuwafohjames asuwafohjames · Answer 2 · 2021-11-23T12:21:57+01:00

The impulse on a 2.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s = 16i

Impulse: This can be defined as change in momentum of a body. The s.i unit of impulse is kgm/s²

The formula of impulse is

I = m(v-u)................ Equation 1

Where, I = impulse exerted on the ball by the wall, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball

From the question,

Given: m = 2.0 kg, v = (-4i+3j) m/s, u = (4i+3j) m/s

Substitute these values into equation 1

I = 2.0[(-4i+3j)-(4i+3j)

I = 2.0(-4i-4i+3j-3j)

I = 2.0(-8i+0j)

I = -16i.

Hence the impulse exerted by the wall on the ball is -16i

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