Problem A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 10 mm 15 o 20 o B C D A

Solving Problems on Your Own Problem mm 15 o 20 o B C D A A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 1. Draw a momentum impulse diagram: The diagram shows the particle, its momentum at t 1 and at t 2, and the impulses of the forces exerted on the particle during the time interval t 1 to t 2.

Solving Problems on Your Own Problem mm 15 o 20 o B C D A A 25- g steel-jacket bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10- mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. 2. Apply the principle of impulse and momentum: The final momentum mv 2 of the particle is obtained by adding its initial momentum mv 1 and the impulse of the forces F acting on the particle during the time interval considered. mv 1 +  F  t = mv 2  F is sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

Problem Solution 10 mm 15 o 20 o B C D A Draw a momentum impulse diagram. += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o Since the bullet leaves a 10-mm scratch and its average speed is 500 m/s, the time of contact  t is:  t = (0.010 m) / (500 m/s) = 2 x s

Problem Solution += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o Apply the principle of impulse and momentum. mv 1 +  F  t = mv 2 (0.025 kg)(600 m/s)cos15 o +F x  2 x s   = (0.025 kg)(400 m/s)cos20 o F x = kN -(0.025 kg)(600 m/s)sin15 o +F y  2 x s  =(0.025 kg)(400 m/s) sin20 o F y = kN + x components: + y components:

Problem Solution += x y F x  t F y  t m v1m v1 x y 15 o x y m v2m v2 20 o F x = kN, F y = kN F = ( kN ) 2 + ( kN ) 2 = 445 kN F = 445 kN 40.1 o