PHYS 2014 NAME: Paper Homework 06 Due: 5pm, 03/16/2018 CWID LAB section: 1. Line

Question

PHYS 2014 NAME: Paper Homework 06 Due: 5pm, 03/16/2018 CWID LAB section: 1. Linear Momentum Conservation Total Points: 10 Two spheres of masses m 50.0 g and M = 85.0 g are suspended by vertical cords as shown in figure below. The sphere with mass m is pulled to the left to a height h = 9.00 cm and then released from rest. After swinging down, it makes an elastic collision with the other sphere which is initially at rest. Two important things that you need to understand for this problem, first, the system is isolated and second, the collision can be assumed to be one-dimensional, since the time of collision is very short (a) After this collision, what heights are reached by the two spheres? (b) After the next collision, which is also elastic, what heights are reached by the two spheres? m M

Explanation / Answer

given

m = 50 g

M = 85 g

hm = 9 cm

a. just after collision, spee dof two masses = u and U

then from conservation of meomentum

mv = MU - mu

now,

from conservation of energy

0.5mv^2 = mgh

v = sqrt(2gh)

hence

m*sqrt(2gh) = MU - mu

U = m(sqrt(2gh ) + u)/M

from conservation of enregy

mgh = 0.5(mu^2 + MU^2)

mgh = 0.5(mu^2 + m^2(sqrt(2gh) + u)^2/M)

2Mgh = (Mu^2 + 2mgh + mu^2 + 2u*msqrt(2gh))

u^2(M + m) + u(2m*sqrt(2gh)) + 2gh(m - M) = 0

hence

0.135u^2 + 0.1328834u - 0.061803 = 0

u = 0.344514 m/s

U = 0.984322 m/s

hence, heights reached by two spheres

h' = u^2/2g = 0.6049434 cm

H' = U^2/2g = 4.9382762 cm

b. after second collision, speeds be v, V

then from conservation of momentum

0.050*sqrt(2*9.81*0.09) = MV - mv = 0.0664417

V = (0.0664417 + 0.05v)/0.085 = 0.7816671 + 0.5882352v

from conservation of energy

0.05*9.81*0.09 = 0.5(0.05v^2 + 0.085V^2)

0.08829 = 0.05v^2 + 0.085(0.61100345522241 + 0.34602065051904*v^2 + 0.91960820580384v )

0.0794117552v^2 + 0.078166v -0.0363547 = 0

hence

v = 0.34451439 m/s

V = 0.98432259 m/s

final heights

h" = 0.6049447753 cm

H" = 4.9382821cm

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