Engineering
Physics
Newtons Second Law of Motion
Projectile Motion
Momentum and Energy Conservation for System of Particles
Question

Three different projectiles, each with the same mass, are fired with speed v at a wall. In case A, the projectile bounces straight back with speed v. In case B, the projectile sticks to the wall. In case C, the projectile crashes through the wall and emerges with half its original speed. These three cases are shown here. Place the impulse exerted by the wall on the projectile in each of these three cases in the correct order.

A > B > C

C > B > A

B > A > C

A > C > B

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Solution
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Impulse (J) equals change in momentum: J = mΔv. All projectiles have mass m and initial velocity +v (toward wall).

Case A (bounces back): Δv = vfinal - vinitial = (-v) - (+v) = -2v. So |JA| = |m(-2v)| = 2mv.

Case B (sticks): Δv = 0 - v = -v. So |JB| = mv.

Case C (emerges at v/2): Δv = (+v/2) - (+v) = -v/2. So |JC| = m(v/2) = 0.5mv.

Therefore, the order of impulse magnitude is A > B > C.

Final Answer: A > B > C