A cart of mass m1 14 kg slides down a frictionless ramp and is made to collide w
ID: 1775655 • Letter: A
Question
A cart of mass m1 14 kg slides down a frictionless ramp and is made to collide with a second cart of mass m2 = 22 kg which then heads into a vertical loop of radius 0.21 m as shown in Figure P7.66 (a) Determine the minimum height h at which cart #1 would need to start from to make sure that cart #2 completes the loop without leaving the track. Assume an elastic collision. Submit Answer Tries o/10 (b) Find the height needed if instead the more massive cart is allowed to slide down the ramp into the smaller cart. Submit Answer Tries 0/10 Post Discussion send FeedbackExplanation / Answer
ELASTIC COLLISION
mass of cart 1 = m1
mass of cart 2 = m2
velocity before collision
velocity of cart 1 = v1i = sqrt(2*g*h)
velocity of cart 2 = v2i = 0
velocity after collision
velocity of car 1 = v1f
velocity of car 2 = v2f
initial momentum before collision
Pi = m1*v1i + m2*v2i
after collision final momentum
Pf = m1*v1f + m2*v2f
from momentum conservation
total momentum is conserved
Pf = Pi
m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2
KEf = 0.5*m1*v1f^2 + 0.5*m2*v2f^2
KEi = KEf
0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)
solving 1&2
we get
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
v2f = ( (22-14)*0 + (2*14*sqrt(2*g*h)))/(14+22)
v2f = (7/9)*sqrt(2*g*h)
for the mass m2 to complete teh circle the speed v2f = sqrt(5*g*r)
sqrt(5*9.8*0.21) = (7/9)*sqrt(2*9.8*h)
height h = 0.867 m <<<<<-----------ANSWER
--------------------------------
if masses are exchanged
v2i = sqrt(2*g*h)
v1i = 0
v1f = ( (m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v1f = (0 + (2*22*sqrt(2*g*h)))/(14+22)
v1f = (11/9)*sqrt(2gh)
for the mass m2 to complete teh circle the speed v1f = sqrt(5*g*r)
sqrt(5*9.8*0.21) = (11/9)*sqrt(2*9.8*h)
height h = 0.35 m <<<<<-----------ANSWER
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