Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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 Feedback

Explanation / 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

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote