A machine gun is fired at a steel plate. Is the average force on the plate from the bullet impact greater if the bullets bounce off or if they are squashed and stick to the plate? Explain.
Solution 14DQ Step 1 of 3: From newton's second law, we know that; net force on the particle is equal to the rate of change of momentum. That is F= dt Where F is the net force and P is the momentum. Step 2 of 3: In the given problem, let P be the initial momentum When the bullets are bounced back from the plate, Change in momentum is, F = dP bounce dt Fbounce= P-(-P)= 2P Similarly when the bullets are squashed from the plate its speed decrease, dP Change in momentum is, Fsquashed dt Fsquashed P-P 1 Similarly when the bullets stick the plate its final speed drops to zero resulting momentum to zero, Change in momentum is, F = dP stick dt Fstick P-0= P
