Project 2 Newtons Laws

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Newtons First Law: A body remains at rest or in motion at a constant speed in a straight line unless acted upon by a force. Second Law: When a body is acted upon by a force, the time rate of change of its momentum equals the force. Momentum, p=m x v Force, F = p ⅆ t ⅆ = ⅆ t ⅆ ( m×v ) m v ⅆ t ⅆ + v m ⅆ t ⅆ If mass is constant, then… F = m v ⅆ t ⅆ = ma Where a is the acceleration of the body Newton’s Third Law: If 2 bodies exert forces on each other, these forces have the same magnitude but opposite directions. In this experiment, we will show that Newton’s 1 st , 2 nd , and 3 rd laws of motion can be used to describe the outcome of each attempt. When an object is dropped from a certain height it has potential energy depending on the height it’s dropped from. The gravitational force will influence the rate of the fall. Newton’s 1 st law states that the object will fall at the same rate if it is not affected by any force. The only forces that will act against the object are wind resistance and impact force. The 2 nd law of motion is exhibited by the force released from the ground on the object at impact. This is known as the impulse force F = m v ⅆ t ⅆ = ma Impulse, I = ∫ 0 t F ⋅ t ⅆ = ∫ u V m ⋅ v ⅆ Impact force is calculated using… F impact = m ( v − u ) Δt In this equation “v” is velocity of the phone at the time of collision and “u” is the velocity when it is dropped The 3 rd law of motion states that the phone will exert force on ground equal to its weight and the ground on the phone but in the opposite direction.

Phone without case: m = 6.2 oz = 0.175 kg 0.02834 * 6.2 oz = 0.175708 kg Time to impact Δt = 0.01 s P E i + k E i = P E f + k E f mgh + 0 = 1 2 mv 2 + 0 Where “v” is velocity of phone before impact. mgh = 0.5 ⋅ m v 2 v = √ 2 ⋅ 9.81 ⋅ 1.5 v = 5.425 m s − 1 Deceleration at impact can be calculated as, v f = v + aΔt Here v f is the velocity of the phone after impact which is 0 and a = deceleration. 0 = 5.425 + 0.01 ⋅ a a =− 542.5 m s − 2 Force of impact, F impact = m ( v − u ) Δt = 0.175 ( 5.425 − 0 ) 0.01 F impact = 94.93 N Phone w/ Silicon case Phone mass: m 1 = 6.2 oz = 0.175 kg

Silicon Case: m 2 = 1.7 oz = 0.048 kg Combined mass = m = m 1 + m 2 =0.175 + 0.048 = 0.223 kg Time to impact Δt = 0.05 s P E i + k E i = P E f + k E f mgh + 0 = 1 2 mv 2 + 0 mgh = 0.5 ⋅ m v 2 v = √ 2 ⋅ 9.81 ⋅ 1.5 v = 5.425 m s − 1 Deceleration of impact 0 = 5.425 + 0.05 ⋅ a a =− 108.5 m s − 2 Force of collision = 0.223 ( 5.425 − 0 ) 0.05 F impact = 24.19 N Hard plastic case and phone Phone mass: 6.2 oz = 0.175 kg Hard plastic case 1.1 oz = 0.031 kg Combined mass = 0.175 + 0.031 = 0.206 kg Time to impact Δt = 0.03 s v = √ 2 ⋅ 9.81 ⋅ 1.5 v = 5.425 m s − 1 Deceleration at impact 0 = 5.425 + 0.03 ⋅ a a =− 180 . 83 m s − 2 Force of collision

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= 0.206 ( 5.425 − 0 ) 0.03 F impact = 37.25 N Rubber case + phone Phone mass: 6.2 oz = 0.175 kg Rubber case 3.2 oz = 0.0907 kg = 0.175 + 0.0907 = 0.2657 kg Time to impact Δt = 0.08 s v = √ 2 ⋅ 9.81 ⋅ 1.5 v = 5.425 m s − 1 0 = 5.425 + 0.08 ⋅ a a =− 67.81 ms − 2 = 0.2657 ( 5.425 − 0 ) 0.0 8 F impact = 18.01 N 1. Velocity before collision remains constant in every case. This means the velocity is independent of mass and collision time only the height the object is dropped from. 2. Acceleration is determined by gravity and since no other forces influence the rate of the fall the acceleration remains constant. Deceleration at impact is determined by collision time. When collision time decreases deceleration increases. 3. Force of Impact is determined by the combined mass and collision time. Since velocity is constant so is force. Impact force is the most significant factor in determining the damage fall and impact have on the phone. From all the examples above the material that absorbed the most force from the impact is the rubber case. This material only generated 18.01 N which was the least of all the materials used in this experiment. My recommendation would be to use the rubber case. The main objective of any phone case is to protect the device from damage. The rubber case is the most effective option in accomplishing this protection.

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