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Step 7: Establish the Heights from which to Drop the Soda Bottle
You will drop the bottle/water mass so that it hits the To achieve a speed of 2 m/s, the bottle must be
lever at different speeds. Since an object in free fall is dropped at 0.20 m.
accelerated by gravity, you need to determine the heights To achieve a speed of 3 m/s, the bottle must be
necessary to drop the bottle to achieve the speeds of 2
m/s, 3 m/s, 4 m/s, 5 m/s, and 6 m/s. Use the
dropped at 9.02 m.
equation Ht=
2
to calculate the height, where Ht is the To achieve a speed of 4 m/s, the bottle must be
29
dropped at 0.80 m.
height, v is the speed (velocity), and g is the gravitational To achieve a speed of 5 m/s, the bottle must be
acceleration of 9.8 m/s2
dropped at
Record these heights in Table B.
To achieve of 6 m/s, the bottle must be
m.
dropped at 1.28
m.
1.56
1.84
2.56
D

    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