A worker give an impulse of 4,1 N*s to a stationary object of 0,21 kg. What is the speed of the object after impact.

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll recall the fact that the impulse represents the change in momentum of an object.

I = F*delta t

F is the force applied to the object over the time interval delta t

But F = m*a => I = m*a*delta t

But a = delta v/delta t => I = m*delta v*delta t/delta t

I = m*delta v (1)

delta v = vf - vi

vf is the final speed of the object and vi is the initial speed of the object.

We'll plug into the equation (1) the given information:

4.1 = 0.21*(Vf - 0).

The initial speed is zero since the object is stationary, at first.

4.1 = 0.21*Vf => vf = 4.1/0.21

vf = 19.52 m/s

Therefore, the speed of the object, after the impact, is of 19.52 m/s.

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question