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

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SUMMARY

The impulsive force during the impact of a mobile phone dropped from a height of 1.5 meters is calculated to be approximately 257 N. The mass of the phone is 0.2 kg, and it bounces to a maximum height of 50 mm after the impact. The contact time with the floor is 0.005 seconds. The initial velocity (v0) was calculated to be 5.42 m/s, and the final velocity (v1) was found to be 0.99 m/s, with the correct application of signs being crucial for accurate results.

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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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