2425- Fall7-BROOKS Activities and Due Dates Chapter 9: Linear Momentum and Colli

Question

2425- Fall7-BROOKS Activities and Due Dates Chapter 9: Linear Momentum and Collisions 7/2017 11:00 PM A 154/100 10/4/2017 05:18 PM Calculator Periodic Table Print 5 of 13 Sapling Learning Map A 0.250-kg lump of dlay is dropped from a height of 1.75 m onto the floor. It sticks to the floor and does not bounce What is the magnitude of the impulse J imparted to the clay by the floor during the impact? Number kg m/s The force exerted by the floor on the clay is plotted as a function of time in the figure to the right. What must have been the maximum force Fmax exerted by the floor on the clay? Number 9.5 ms Previous Check Answer O Next Ex nt

Explanation / Answer

Given,

m = 0.25 kg ; h = 1.75 m ;

We know that the impulse is rate of change of momentum.

velocity while its hits,

v = sqrt (2 g h) [from conservation of energy]

v = sqrt(2 x 9.8 x 1.75) = 5.86 m/s

momentum = mv = 0.25 x 5.86 = 1.47 kg-m/s

Hence, J = 1.47 kg-m/s

We know that,

J = F(max) t = m v

Fmax = mv/t

Fmax = 1.47/9.5 x 10^-3 = 150 N

Hence, Fmax = 150 N

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