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**Perfectly Inelastic collisions - by Lauren Bly, 2001**

**1. A 1500 kg car traveling at 15.0 m/s to the south
collides with a 4500 kg truck that is initially at rest at a stoplight.
The car and truck stick together and move together after the collision.
What is the final velocity of the two vehicle mass? **

Here's what you know: The car has mass 1500 kg, and initial velocity 15.0 m/s. The truck has mass 4500 kg and initial velocity 0 m/s. The final mass is 6000 kg (4500 kg plus 1500 kg, because they stick together) and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the car plus p of the truck equals the p of them together. So, 1500 x 15 + 4500 x 0 = 6000 x v final. 22500 = 6000 v, and

v = 3.8 m/s. The velocity is positive, and the initial velocity of the car was positive, so the two vehicles together are traveling to the south just like the car was initially.

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**2. A grocery shopper tosses a 9.0 kg bag of rice into a
stationary 18.0 kg shopping cart. The bag hits the cart with a horizontal
speed of 5.5 m/s toward the front of the cart. What is the final speed of
the cart and bag? **

Here's what you know: The rice has mass of 9.0 kg and initial velocity of 5.5 m/s. The cart has a mass of 18.0 kg and an initial speed of 0 m/s. Together, the bag and cart have a mass of 27 kg and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the bag before plus the p of the cart before equals the p of the cart and bag together. So 9 x 5.5 + 18.0 x 0 = 27 x v final. 49.5 + 0 = 27 v, and

v = 1.83 m/s.

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**3. A 1.50x10^4 kg railroad car moving at 7.00 m/s to the
north collides with and sticks to another railroad car of the same mass that is
moving in the same direction at 1.50 m/s. What is the velocity of the
joined cars after the collision?**

Here's what you know: The first car has mass of 1.5 x x10^4 kg and initial velocity of 7.00 m/s. The second car has a mass of 1.5 x x10^4 kg and an initial speed of 1.5 m/s. Together, the two cars have a mass of 3 x 10^4 kg and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the first car before plus the p of the second car before equals the p of the two cars together. So 1.5 x x10^4 x 7.00 + 1.5 x x10^4 x 1.5 = 3 x x10^4 x v final. 1.27500 x 10^5 = 3 x x10^4 v, and

v = 4.0 m/s to the north. All the velocities are positive, so they al are in the same direction, which is north.

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**4. A dry cleaner throws a 22 kg bag of laundry onto a
stationary 9.0 kg cart. The car and laundry bag begin moving at 3.0 m/s to
the right. Find the velocity of the laundry bag before the collision.**

Here's what you know: The laundry bag has mass of 22 kg and an unknown velocity. The cart has a mass of 9.0 kg and an initial velocity of 0 m/s. Together, the bag and cart have a mass of 31 kg and a final velocity of 3.0 m/s. Using the formula p = mv, and the concept that p before equals p after, the p of the bag before plus the p of the cart before equals the p of the cart and bag together. So 22 x v + 9.0 x 0 = 31 x 3.0 . 22 v + 0 = 93, and

v = 4.2 m/s, to the right.

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**5. A 47.4 kg student runs down the sidewalk and jumps
with a horizontal speed of 4.20 m/s onto a stationary skateboard. the
student and the skateboard move down the sidewalk with a speed of 3.95
m/s. Find the following:**

** a. the mass of the skateboard**

** b. how fast the student would have to jump to have a
final speed of 5.00 m/s. **

a. Here's what you know: The student has mass of 47.4 kg and initial velocity of 4.20 m/s. The skateboard has an unknown mass and an initial velocity of 0 m/s. Together, the student and skateboard have a mass of (47.4 + m) and a final velocity of 3.95 m/s. Using the formula p = mv, and the concept that p before equals p after, the p of the student before plus the p of the skateboard before equals the p of the student and skateboard together. So 47.4 x 4.20 + m + 0 = (m + 47.4) x 3.95. (199 + 0) / 3.95 = 47.4 m, and

m = 3.0 kg.b. Here's what you know: The student has mass of 47.4 kg and initial velocity unknown. The skateboard has a mass of 3.0 kg and an initial velocity of 0 m/s. Together, the student and skateboard have a mass of 50.4 kg and a final velocity of 5.00 m/s. Using the formula p = mv, and the concept that p before equals p after, the p of the student before plus the p of the skateboard before equals the p of the student and skateboard together. So 47.4 x v + .248 + 0 = 50.4 x 5.0. 47.4 v + 0 = 252, and

v = 5.32 m/s.

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