From the half-life for 14c decay, 5715 yr, calculate the age of the shroud.

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"The cloth shroud from around a mummy is found to have a 14C activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for 14C decay, 5715 yr, calculate the age of the shroud. (Radioactive decay follows the first-order kinetics.) "

Solution:
[A]₀ = 16.3 dpm/g                    
[A]∨t = 9.7 dpm/g
t ∨1/2 = 5715 yr.
t=?

[A]₀ / [A]∨t = 16.3/9.7 = 2 ∧t/t ∨1/2
t/t∨1/2 = log 1.168/ log 2 = 0.225 / 0.301
           = 0.747
5715 years × 0.747 = 4270 years

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s