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a body of mass 10kg moving with a velocity 4m/s strikes a vertical wall horizontally and returns back with same velocity. find the impulse exerted on the wall if the time of contact is 0.02s​

Respuesta :

abidemiokin

Answer:

1.96Ns

Explanation:

Impulse = Ft = m(v-u)

Impulse = Force * time

Impulse = mgt

Impulse = 10 * 9.8 * 0.02

Impulse = 98*0.02

Impulse = 1.98Ns

Hence the impulse exerted is 1.96Ns

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