5. Three carts of masses m 1 = 4.50 kg, m 2 = 7.50 kg, and m 3 = 3.00 kg move on

Question

5. Three carts of masses m1 = 4.50 kg, m2 = 7.50 kg, and m3 = 3.00 kg move on a frictionless, horizontal track with speeds of v1 = 6.50 m/s to the right, v2 = 3.00 m/s to the right, and v3 = 5.00 m/s to the left, as shown below. Velcro couplers make the carts stick together after colliding.

(a) Find the final velocity of the train of three carts.

magnitude

= .... m/s

direction

  -right or left

(b) Does your answer require that all the carts collide and stick together at the same moment? Yes, or No

What if they collide in a different order?

magnitude

= .... m/s

direction

  -right or left

Explanation / Answer

(a)

initial momentum Pi = m1*v1 + m2*v2 + m3*v3

Pi = (4.5*6.5) + (7.5*3) - (3*5)

Pi = 36.75 kg m/s

after coupling

final moentum Pf = (m1+5m2_m3)*V

from moemntum conservation

Pf = pi

(4.5+7.5+3)*v = 36.75

v = 2.45 m/s

direction right

==========================

(b)

NO

(c)

If m1 stick to m2

(m1*v1) + (m2*v2) = (m1+m2)*v12

(4.5*6.5) + (7.5*3) = (4.5+7.5)*v12

v12 = 4.3125 m/s

m3 sticks to m1&m2

(m1+m2)*v12 - (m3*v3) = (m1+m2+m3)*v

((4.5+7.5)*4.3125 ) - (3*5) = (4.5+7.5+3)*v

v = 2.45 m/s

If m1 stick to m2

(m3*v1) + (m2*v2) = (m3+m2)*v12

-(3*5) + (7.5*3) = (3+7.5)*v23

v23 = 0.714 m/s

m1 sticks to m3&m2

(m3+m2)*v12 + (m1*v1) = (m1+m2+m3)*v

((3+7.5)*0.714 ) + (4.5*6.5) = (4.5+7.5+3)*v

v = 2.45 m/s

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