Question

a 2.0 kg ball is dropped from a height of 20 m onto a soft surface and rebounds to a height of 5.0 m . what is the magnitude of the impulse exerted on the ball by the floor?

Answer 1

Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.

What is the impulse of the floor on the ball?

Using the equation of motion to determine the velocity at the end of the fall

• v^2 = u^2 + 2gh

Where v is velocity at the end of fall

u is initial velocity = 0

g is acceleration due to gravity = 9.81 m/s^2

h is height = 20

• Taking downward velocity as negative and up as positive

v^2 = 0 + 2 (9.81)(20)

v^2 = 392.4

v = - 19.8 m/s

The velocity, v after bouncing is calculated also:

u = 0

g = 9.81 m/s^2

h = 5.0 m

v^2 = 0 + 2(9.81)(5)

v^2 = 98.1

v = 9.904 m/s

• Impulse = change in momentum
• Impulse = m(v- u)

Impulse = 2.0 × (9.9 -(-19.8)

Impulse = 59.4 Ns

Therefore, the impulse of the floor on the ball is 59.4 Ns.

Learn more about impulse at: brainly.com/question/904448