A m1 = 7.3 g, bullet is shot through a m2 = 1.32 kg, wood block suspended on a s

Question

A m1 = 7.3 g, bullet is shot through a m2 = 1.32 kg, wood block suspended on a string 2.1 m, long. The center of mass of the block rises a distance of 2.2 cm . The initial speed of the bullet is v1i = 451 m/s.

A) Find the speed of the block immediately after the bullet leaves it.

B) Find the speed of the bullet immediately after the collision.

Explanation / Answer

A) we have velocity of wood block =>h=v^2/2g 0.022=v^2/(2*9.8) =>v= sqrt(2*9.8*0.022) = 0.656 m/s speed of the block immediately after the bullet leaves=0.656 m/s B) we have m1u1+m2u2=m1v1+m2v2 =>0.0073*451 + 0 = 0.0073*v1 + 1.32*0.656 =>v1 = (0.0073*451-1.32*0.656)/0.0073 = 332.381 m/s => speed of the bullet immediately after the collision is 332.381 m/s

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