How Is the Impulsive Force Calculated During a Mobile Phone's Impact?

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The discussion focuses on calculating the impulsive force experienced by a mobile phone dropped from a height of 1.5 meters, with a mass of 0.2 kg. After bouncing to a height of 50 mm and being in contact with the floor for 0.005 seconds, the average impulsive force is estimated to be around 257 N. The user initially calculated the velocities using conservation of energy, arriving at values of 5.42 m/s for v0 and 0.99 m/s for v1, but did not account for the sign of the velocities, leading to an incorrect force calculation of approximately 177 N. The correction emphasizes the importance of considering direction in velocity calculations. Accurate calculations of impulsive force are crucial in understanding the impact dynamics of objects.
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Homework Statement


Mobile phone is dropped onto a hard floor from a height of 1.5m. The mass of the mobile phone is 0.2 kg. After the impact, the phone bounces to a maximum height of 50 mm. The phone is estimated to be in contact with the floor for 0.005s. Which is of the following is the closest to the average value of the impulsive force during the impact?

Homework Equations


F dt = dP
dP = mv1 - mv0


The Attempt at a Solution


so what i basically tried to do was calculate v1 and v0 then can find F but the answer is supposed to be 257 N where as I am getting ~177N
so i calculated the velocities using conservation of energy and found v0 to be 5.42 m/s and v1 to be 0.99 m/s and then rearrange to solve for F.
 
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You did not take the sign of the velocities into account.

ehild
 
oh wow I am so silly... thankyou so much!
 

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