Question

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor

Answer 1

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

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

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

Answer 2

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

• m = 2.4 kg

Final velocity,

• v = 2.5 m/s

Initial velocity,

• u = -1.5 m/s

By using Newton's 2nd law of motion, we get

→ Impulse, $I = m(v-u)$

By substituting the values, we get

                     $= 2.4[2.5-(1.5)]$

                     $= 2.4(2.5+1.5)$

                     $= 2.4\times 4$

                     $= 9.6 \ Ns$

Thus the above answer is right.    

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