Practice 6G: | 1 | 2 | 3 | 4 | Go up
Elastic collisions- by Lauren Bly, 2001

1. A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 18.0 cm/s  After the collision, the first marble moves to the left at 18.0 cm/s.

a. Find the velocity of the second marble after the collision.

b. Verify your answer by calculating the total kinetic energy before and after the collision.

a. Here's what you know: The right marble has a mass of 0.015 kg and an initial velocity of 22.5 cm/s (0.225 m/s). The left marble has a mass of 0.015 kg and an initial velocity of -18.0 cm/s (-0.18 m/s).  After the collision, the marbles have the same masses, but the right marble has a new velocity of -18 cm/s (-0.18 m/s), and the left marble’s new velocity is unknown. Using the formula p = mv, and the concept that p before equals p after, the p of the right marble before plus the p of the left marble before equals the p of the right marble after plus the left marble after.  So 0.015 x 0.225 + 0.015 x -0.18 = 0.015 x -0.18 + 0.015 x v final. 6.75 x 10 -4 = 0.015 x v + -0.0027, and v = .225 m/s, or 22.5 cm/s.

b. Here’s what you know: The kinetic energy measured before and after will be the same, because this happens in elastic collisions and this is an elastic collision.  The before and after kinetic energies are calculated with  KE =1/2mv2 .  The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before, and add the values of the two objects after to get the total after.  The value of KE before is (0.5)(0.015)(.225)2 + (0.5)(0.015)(18) 2.  Therefore, the KE before is 6.2 x 10-4 J.  The KE afterward is (0.5)(0.015)(.18) 2 + (0.5)(0.015)(0.225)2, which equals 6.2 x 10-4 J.   6.2 x 10-4 J is equal to 6.2 x 10-4 J, and so the KE is conserved.

2. a 16.0 kg canoe moving to the left at 12 m/s makes an elastic head-on collision with a 4.0 kg raft moving to the right at 6.0 m/s.  After the collision, the raft moves to the left at 22.7 m/s.  disregard and effects of the water.

a. Find the velocity of the canoe after the collision.

b. Verify your answer by calculating the total kinetic energy before and after the collision.

a. Here's what you know: The canoe has a mass of 16 kg and an initial velocity of 12 m/s. The raft has a mass of 4.0 kg and an initial velocity of -6 m/s.  After the collision, the canoe keep their same masses, but the raft has a new velocity of  22.7 m/s, and the canoe’s new velocity is unknown. Using the formula p = mv, and the concept that p before equals p after, the p of the canoe before plus the p of the raft before equals the p of the canoe after plus the raft after.  So 16 x 12 + 4 x -6 = 16 x v final + 4 x 22.7. 168 = 16 x v + 90.8, and v = 4.83 m/s.

b. Here’s what you know: The kinetic energy measured before and after will be the same, because this happens in elastic collisions and this is an elastic collision.  The before and after kinetic energies are calculated with  KE =1/2mv2 .  The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before, and add the values of the two objects after and get the total after.  The value of KE before is (0.5)(16)(12)2 + (0.5)(4)(6) 2.  Therefore, the KE before is 1224 J.  The KE afterward is (0.5)(16)(4.83) 2 + (0.5)(4)(22.7)2, which equals 1217 J.   1224 J is about equal to 1217 J, and so the KE is conserved.

3. A 4.0 kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially at rest.  The first ball stops after the collision.

a. Find the velocity of the second ball after the collision.

b. Verify your answer by calculating the total kinetic energy before and after the collision.

a. Here's what you know: The first bowling ball has a mass of 8.0 kg and an initial velocity of 8.0 m/s. The second bowling ball has a mass of 4.0 kg and an initial velocity of 0 m/s.  After the collision, the bowling balls keep their same masses, but the first ball has a new velocity of  0 m/s, and the second ball's new velocity is unknown. Using the formula p = mv, and the concept that p before equals p after, the p of the first ball before plus the p of the second ball before equals the p of the fist ball after plus the p of the second ball after.  So 8 x 4 + 4 x 0 = 8 x 0 + 4 x v final. 32 = 0 + 4v, and v = 8 m/s.  The velocity is positive, so the second ball is now moving to the right.

b. Here’s what you know: The kinetic energy measured before and after will be the same, because this happens in elastic collisions and this is an elastic collision.  The before and after kinetic energies are calculated with  KE =1/2mv2 .  The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before, and add the values of the two objects after to get the total after.  The value of KE before is (0.5)(4)(8)2 + (0.5)(4)(0) 2.  Therefore, the KE before is 130 J.  The KE afterward is (0.5)(4)(0) 2 + (0.5)(4)(8)2, which equals 130 J.   130 J is equal to 130 J, and so the KE is conserved.