Question

a 15 g ball is thrown at a barricade. the ball will experience the most impulse if
it comes to a stop when it hits the barricade and falls directly to the floor.
it explodes into pieces and falls to the floor a short distance in front of the barricade
it bounces off the barricade and travels backward half the way back to where it was thrown from.
it knocks over the barricade and rolls slightly forward.

Explanation:

Step1: Recall the impulse formula

Impulse \(J=\Delta p = m\Delta v\), where \(m\) is mass and \(\Delta v\) is the change in velocity.

Step2: Analyze each option

• Option 1: Let initial velocity be \(v\). Final velocity \(v_f = 0\). \(\Delta v=0 - v=-v\)
• Option 2: The pieces still have some forward - component of velocity (even if reduced). Let initial velocity be \(v\). Final velocity \(v_f>0\) (but \(v_f < v\)). \(\Delta v=v_f - v\) (less in magnitude than in option 3)
• Option 3: Let initial velocity be \(v\) (towards the barricade). Final velocity \(v_f=-\frac{v}{2}\) (backward). \(\Delta v=-\frac{v}{2}-v=-\frac{3v}{2}\)
• Option 4: The ball still has a forward - component of velocity (even if reduced). Let initial velocity be \(v\). Final velocity \(v_f>0\) (but \(v_f < v\)). \(\Delta v=v_f - v\) (less in magnitude than in option 3)

Since the mass \(m\) is the same for all cases (\(m = 15g\)), the largest \(\vert\Delta v\vert\) gives the largest impulse.

Answer:

it bounces off the barricade and travels backward half the way back to where it was thrown from.