A bungee jumper (m = 71.00 kg) tied to a 31.00 m cord leaps off a 61.00 m tall b

Question

A bungee jumper (m = 71.00 kg) tied to a 31.00 m cord leaps off a 61.00 m tall bridge. He falls to 7.00 m above the water before the bungee cord pulls him back up. What size impulse is exerted on the bungee jumper while t it cord stretches? (in kg*m/s) 3.16 times 10^2 4.21 times 10^2 5.60 times 10^2 7.44 times 10^3 9.90 times 10^2 1.32 times 10^3 175 times 10^3 2.33 times 10^3 An MSU linebacker of mass 124.0 kg sacks a UM quarterback of mass 75.0 kg. Just after they collide, they are momentarily stuck together, and both are moving at a speed of 2.90 m/s. If the quarterback was at rest just before he was sacked, how fast was the linebacker moving just before the collision? (in m/s) 1.53 2.21 3 21 4.65 6.75 9.79 1.42 times 10^1 2.06 times 10^1 An irregular shaped object of mass 4.40 kg has a moment of inertia 14.476 kgm^2 about an axis through a point 1.30 m from its center of mass. Calculate the moment of inertia about Parallel axis through its center of mass. (in kg*2) 6.02 7.04 8.24 9.64 1.13 times 10^1 1.32 times 10^1 1.54 times 10^1 1.81 times 10^1

Explanation / Answer

a)

distance it falls while stretching the cord = 61 - 31 - 7 = 23 m

Distance it has fallen when it just starts to stretch, d = 31 m

So, speed it has at this point , v = sqrt(2gd) = sqrt(2*9.8*31) = 24.6 m/s

Final speed after it finall comes to rest momentarily = 0

So, impulsle imparted = m*v = 71*24.6 = 1746 kg.m2 = 1.75*10^3 kg.m2

b)

Conserving momentum,

124*v + 75*0 = (124+75)*2.9

So, v = 4.65 m/s

c)

Using parallel axis theorem,

I = Icm + M*d^2

Icm = moment of inertia through the center of mass

d = distance from the center of mass

So, 14.476 = Icm + 4.4*1.3^2

So, Icm = 7.04 kg.m2

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