Two blocks of masses m1 = 1.90 kg and m2 = 5.18 kg are each released from rest a

Question

Two blocks of masses m1 = 1.90 kg and m2 = 5.18 kg are each released from rest at a height of h = 5.15 m on a frictionless track, as shown in the figure, and undergo an elastic head-on collision. m1 from the right and m2 from the left. Determine the velocity of the m2 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.) Determine the velocity of the m1 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.) Determine the maximum height to which m2 rises after the collision. Determine the maximum height to which m1 rises after the collision.

Explanation / Answer

Find the velocities before the collision by conservation of energy

mgh = .5mv^2

v = sqrt(2gh)

v = sqrt(2)(9.8)(5.15) = 10 m/s

Now using conservation of momentum

(1.9)(10) + (5.18)(-10) = (1.9)(v) + (5.18)(u)   I will use u and v to keep them separate

-32.8 = 1.9v + 5.18u

v = (-32.8 - 5.18u)/1.9 = -17.3 -2.73u

Using conservation of energy

.5(1.9)(10^2) + (.5)(5.18)(10^2) = (.5)(1.9)(v^2) + (.5)(5.18)(u^2)

95 + 259 = .95(-17.3 - 2.73u)^2 + 2.59u^2

354 = 284.33 + 89.74u + 9.67u^2

In standard for,

9.67u^2 + 89.74u -69.67 = 0

Find the roots using the quadratic equation

Roots are .72 m/s and -10 m.s

We can eliminate the 10 because it can not rebound with the same initial speed it had, thus the speed after the collision for the 5.18 kg mass is .72 m/s

For the first mass v = -17.3 - 2.73(.72) = -19.3 m/s

So for mass 1 the final speed is -19.3 m/s

For mass 2 the final speed is .72 m/s

Part B)

For the heights

Apply h = v^2/2g

h for mass 1 = 19.3^2/(2)(9.8) = 18.9 m

h for mass 2 = (.72^2)/(2)(9.8) = .026 m (2.6 cm)

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