Print ! Calculator -d Periodic Table Question 8 of 14 sapling leaming Two skydiv

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Print ! Calculator -d Periodic Table Question 8 of 14 sapling leaming Two skydivers are holding on to each other while falling straight down at a common terminal speed of 59.50 m/s Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.80 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis) = 4.930 m / s , = 4.250 m/s r,= 59.50 m / s r s What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.20 kg immediately after separation? Number 1y m s 2.x Number m/s 2,5 What is the change in kinetic energy of the system? Number Joules

Explanation / Answer

m1 =83.8 kg, m2 =52.2kg

u =59.5 m/s along z-direction

ux =0 ,uy =0

v1x = 4.93 m/s

From conservation of momentum along x-direction

(m1+m2)ux = m1v1x+m2v2x

0 = m1v1x+m2v2x

v2x = -(83.8*4.93)/52.2

v2x = - 7.9 m/s

v1y =4.25 m/s

From conservation of momentum along y-direction

(m1+m2)uy = m1v1y+m2v2y

0 = m1v1y+m2v2y

v2y = - (83.8*4.25)/52.2

v2y = - 6.8 m/s

v1z =uz = 59.50 m/s

From conservation of momentum along y-direction

(m1+m2)uz = m1v1z+m2v2z

v2z = 59.5 m/s

initial kinetic energy

Ki = (1/2)(m1+m2)u^2 = (1/2) (83.8+52.2)(59.5*59.5)

Ki = 240737 J

v1 = [(v1x)^2 +(v1y)^2+(v1z)^2]^1/2

v1= [(4.93)^2+(4.25)^2+(59.5)^2]^1/2

v1 =59.85 m/s

v2 = [(v2x)^2 +(v2y)^2+(v2z)^2]^1/2

v2= [(-7.9)^2+(-6.8)^2+(59.5)^2]^1/2

v2 =60.41 m/s

Final kinetic energy

Kf = (1/2)m1v1^2 +(1/2)m2v2^2

= [(1/2)(83.8)(59.85*59.85)]+[(1/2)(52.2)(60.41)(60.1)]

=150086.7+ 95248.5

Kf =245335.2J

Change in kinetic energy = Kf-Ki

=245335.2 -240737

=4598.2 J

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