As shown in the figure below, two blocks ( m 1 and m 2 ) are each released from

Question

As shown in the figure below, two blocks (m1 and m2) are each released from rest at a height of h = 4.48 m on a frictionless track and when they meet on the horizontal section of the track they undergo an elastic collision. If m1 = 2.50 kg and m2 = 4.65 kg, determine the maximum heights to which they rise after the collision. Use the coordinate system shown in the figure.

Y1 Final = _________

Y2 Final = _________

The problem may be divided into three steps. Step one, the blocks slide down the curve and onto the horizontal section of the track. What physical quantities are conserved for this step of the problem? See if you can write an appropriate conservation statement for this step that will allow you to determine the x component of the velocity of the blocks on the horizontal section of track and going into the collision.

Step two, the blocks undergo an elastic collision on the horizontal section of the track. What physical quantities are conserved for this step of the problem? See if you can write appropriate conservation statements for this step of the problem that will allow you to determine the x component of the velocity of the blocks after the collision.

Step three, the blocks slide back up the curved section of the track. How is this step similar to and different from the first step? What physical quantities are conserved for this step of the problem? See if you can write an appropriate conservation statement for this step that will allow you to determine the final height the blocks reach as they travel back up the curved section of the track. m

Explanation / Answer

From Conservation of Enenrgy we have

v = 2gh = (2 * 9.8m/s² * 4.48m) = 9.37 m/s
Now, momentum: initial p = final p
2.5kg * 9.37m/s - 4.65kg * 9.37m/s = 2.5kg*u + 4.65kg*v
where u and v the velocities of m1 and m2, respectively, after impact.

Then
-20.145 = 2.5u + 4.65v

From Conservation of Energy, for an elastic, linear collision, the
relative velocity of approach = relative velocity of separation, or
18.74 m/s = v - u, or v = u + 18.74
Plug that into the momentum equation:
-20.145 = 2.5u + 4.65(u + 18.74) = 7.15u + 87.141
u = -107.29/7.15 = -15.00 m/s
v = 18.74m/s + u = 3.73 m/s

hence, the maximum height is
For m1: h = u² / 2g = (15m/s)² / 19.6m/s² = 11.48 m
For m2: h = (3.73m/s)² / 19.6m/s² = 0.710 m

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