A red cart full of wedding dresses collides with a blue cart full of tuxedos in

Question

A red cart full of wedding dresses collides with a blue cart full of tuxedos in an elastic collision. Ignore friction. Use the information in the drawing. What are the velocities of the red and blue carts after the collision? How far up the hill, d, will the blue cart travel? By how much, x, will the red cart compress the spring before coming to a complete stop. Optional Challenge: Without friction, the carts will collide again with the same speeds you computed in part. Assume they collide on the horizontal surface. Show that the final speeds after this second collision are equal to the initial speeds of the carts! (You will need to change reference frames to do this.) m_B = 50 m_R= 30 kg V_Bi = 0m/s v_Ri = -10.0 m/s

Explanation / Answer

a) Just After the collision

VRf = (mR - mB)*vRi/(mR - mB)

= (30 - 50)*(-10)/(30 + 50)

= 2.5 m/s

vBf = 2*mR*vRi/(mR + mB)

= 2*30*(-10)/(30 + 50)

= -7.5 m/s

b) let h is the vertical height travelled blue cart after the collision.

h = vBf^2/(2*g)

= 7.5^2/(2*9.8)

= 2.87 m

d = h/sin(30)

= 2.87/sin(30)

= 5.74 m

c) Apply conservation of energy

(1/2)*k*x^2 = (1/2)*mR*VRf^2

x = sqrt(mR*VRf^2/k)

= sqrt(30*2.5^2/1000)

= 0.433 m

d) When they collide again, just before the collision,

VRi = -2.5 m/s

VBi = 7.5 m/s

when we apply conservation of momentum and conservation of kinetic energy

after the collision, again we get

VRf = 2.5 m/s

VBf = -7.5 m/s

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