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Problem 13.195

D. 10 mm. 20 o. C. A. B. 15 o. Problem 13.195. 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

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Problem 13.195

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  1. D 10 mm 20o C A B 15o Problem 13.195 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. Problem 13.195 D 10 mm 20o C A B 15o Solving Problems on Your Own 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 t1 and at t2, and the impulses of the forces exerted on the particle during the time interval t1 to t2.

  3. Problem 13.195 D 10 mm 20o C A B 15o Solving Problems on Your Own 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 mv2 of the particle is obtained by adding its initial momentum mv1 and the impulse of the forces F acting on the particle during the time interval considered. mv1 +S FDt = mv2 S Fis sum of the impulsive forces (the forces that are large enough to produce a definite change in momentum).

  4. Problem 13.195 Solution y y y mv1 mv2 x x x + = 20o 15o FxD t FyD t D 10 mm 20o Draw a momentum impulse diagram. C A B 15o Since the bullet leaves a 10-mm scratch and its average speed is 500 m/s, the time of contact D t is: D t = (0.010 m) / (500 m/s) = 2x10-5 s

  5. Problem 13.195 Solution y y y mv1 mv2 x x x + = 20o 15o FxD t FyD t + xcomponents: + ycomponents: Apply the principle of impulse and momentum. mv1 +S FDt = mv2 (0.025 kg)(600 m/s)cos15o+Fx(2x10-5s)=(0.025 kg)(400 m/s)cos20o Fx = - 254.6 kN -(0.025 kg)(600 m/s)sin15o+Fy(2x10-5s)=(0.025 kg)(400 m/s) sin20o Fy = 365.1 kN

  6. Problem 13.195 Solution y y y mv1 mv2 x x x + = 20o 15o FxD t FyD t Fx = - 254.6 kN, Fy = 365.1 kN F = ( -254.6 kN )2 + ( 365.12 kN )2 = 445 kN F = 445 kN 40.1o

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