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Mr. Macdonald and Mr. Gong are testing a new ride they designed for the santa cr

ID: 1500256 • Letter: M

Question

Mr. Macdonald and Mr. Gong are testing a new ride they designed for the santa cruz boardwalk. Mr. Gong's cart (m_1 = 320 kg) starts from rest at the top, and slides down the frictionless incline plane (theta_0 = 37 degrees, and horizontal distance b - 6.7 m). His cart continoues on the frictionless surface (eith the same speed it had at the end of the incline) and collides with Mr. Macdonald's stationary cart (m_2 = 270 kg). The two carts lock together and shoot off the vertical wall (h = 1.5m) onto a thick foam pad; what fun! Find X_f (measured from the vertical wall) where the cars land F

Explanation / Answer

As mass m1 starts off from the top of the incline, it has energy stored in the form of Potential energy. When is slides down to the level of m2, it attains some velocity and the potential energy changes to kinetic energy.

We will use principle of conservation of energy to determine the velocity of the block m1, right before it hits m2.

That is, m1gH = 0.5 m1 v2

or, V = sqrt(2gH) = sqrt(2 x 9.81 x 6.7 Sin37) = 8.8944 m/s

Now, as the mass m1 hits the mass m2, they stick together and move forward. Here will use the principle of conservation of linear momentum to find the speed of the combined mass.

That is, m1 v = (m1 + m2)V

or, V = 320 x 8.8944 / (320 + 270) = 4.824

Now the combined mass is thrown off the wall with the speed of 4.824 m/s

While falling through, the combined mass will have no horizontal acceleration, while it will have gravitational acceleration in vertical direction. Also, the masses will have no initial velocity in the vertical direction.

We will determine the time required for the mass to fall verical by h and then use that time to find the horizontal distance covered by the time it hits the floor.

For vertical fall:

S = 0.5 gt2

or, t = sqrt(2h/g) = sqrt(2 x 9.81 / 1.5) = 3.617 seconds

Therefore the horizontal distance covered by the time it hits the floor is horizontal velocity x time

Hence, Xf = 4.824 x 3.617 = 17.448408 metres

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