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Kazeer [188]
2 years ago
12

An excited 92 kg football player celebrates a touchdown by carelessly running straight into the goalpost at 9.4 m/s. He bounces

straight backward at 6.2 m/s. What was the change in velocity? What was the impulse of the player?
Physics
1 answer:
yKpoI14uk [10]2 years ago
7 0

Answer:

i. 15.6 m/s

ii. I = 1.44 KNs

Explanation:

The impulse, I, on a body is the product of force applied on it and the time it acts.

i.e I = F x t

Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.

i. mass, m, of the player = 92 kg

initial velocity of the player, u = 9.4 m/s

final velocity of the player, v = 6.2 m/s

Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.

So that,

change in velocity of the player = final velocity - initial velocity

                                          = 6.2 - (-9.4)

                                         = 6.2 + 9.4

                                         = 15.6 m/s

change in velocity of the player is 15.6 m/s

ii. Impulse, I = m(v - u)

                    = 92 x 15.6

                    = 1435.2

Impulse on the player is 1.44 KNs.

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