Question

3. A5 g bullet a is fired with velocity v. = 500 m/s into a 1-kg wood block B suspended by a cord of length L=2 m. Find the maximum angle the pendulum reaches, 0, and the tension on the cord immediately after impact. (30 point) ө To B

Answer 1

0 Mass of bullet, mo = 59 - 5x100-kg Velocity of bullet, Vo = 500 m/s Mass of block B, MB = 1 kg Let the velocity of combined bullet and block B just after impact bew'. momentum principle, momentum of bullet and block B combined must be equal just before and just aftien the impact. ie, mo vot moxo mov + mov 5x10-3x 500 -3 5x103 xV+ V 2.5 v (5 x 10-3 +1) = 2.488 m/s V = 2m (2-4) B At the maximum augle the velocity of combined bullet and

and they will be at a block B will be zero height from the initial position in the vertical direction. By energy conservation total change in kinetic energy is equal to total change in potential energo. ie, 2 (Motma) xv² - 3 (Mot mb)xo? =(Mot maxgxh final vebrity is gero. acceleration due to gravity = 9-81m/s2 į(5x10-3+1Dx2.4882 – 15x10°3+1)*9-81xu :.h=0.315m From the figure, cose = (2:1)-(2-0,35) = 0.842 O= 32:616° Let us draw the free body diagram of the combined bollet and block B aftesi impact showing all the forces. T tension V (Mox MB) xg

MB) xg ::T (mot ma T = =(5x10-2 +1) *9.81 tension, T 9.859N

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