Practice 6E: | 1 | 2 |
3 | 4 | 5 | Go up
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.
(Table of contents)
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.
(Table of contents)
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.
(Table of contents)
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.
(Table of contents)
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.
(Table of contents)