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Elastic Collision worksheet

Grades 11–12 · CCSS HSA.REI.C.6

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Type of collision: Perfectly inelastic (stick together)
- Mass of object 1 (m₁): 3 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Show velocities in: Meters per second (m/s)
Find: Common final velocity
Answer
2.
Problem 2.
- Type of collision: Perfectly inelastic (stick together)
- Mass of object 1 (m₁): 1 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 2 kg
- Initial velocity of object 2 (u₂): -2 m/s
- Show velocities in: Feet per second (ft/s)
Find: Common final velocity
Answer
3.
Problem 3.
- Type of collision: Elastic (no energy lost)
- Mass of object 1 (m₁): 2 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -2 m/s
- Show velocities in: Feet per second (ft/s)
Find: Final velocity of object 1 (v₁)
Answer
4.
Problem 4.
- Type of collision: Elastic (no energy lost)
- Mass of object 1 (m₁): 3 kg
- Initial velocity of object 1 (u₁): 4 m/s
- Mass of object 2 (m₂): 2 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Show velocities in: Miles per hour (mph)
Find: Final velocity of object 1 (v₁)
Answer
5.
Problem 5.
- Type of collision: Partly elastic (restitution e)
- Mass of object 1 (m₁): 2 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Coefficient of restitution (e): 0.4
- Show velocities in: Meters per second (m/s)
Find: Final velocity of object 1 (v₁)
Answer
6.
Problem 6.
- Type of collision: Elastic (no energy lost)
- Mass of object 1 (m₁): 1 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Show velocities in: Meters per second (m/s)
Find: Final velocity of object 1 (v₁)
Answer
7.
Problem 7.
- Type of collision: Elastic (no energy lost)
- Mass of object 1 (m₁): 3 kg
- Initial velocity of object 1 (u₁): 4 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Show velocities in: Feet per second (ft/s)
Find: Final velocity of object 1 (v₁)
Answer
8.
Problem 8.
- Type of collision: Partly elastic (restitution e)
- Mass of object 1 (m₁): 2 kg
- Initial velocity of object 1 (u₁): 2 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Coefficient of restitution (e): 0.5
- Show velocities in: Feet per second (ft/s)
Find: Final velocity of object 1 (v₁)
Answer
9.
Problem 9.
- Type of collision: Perfectly inelastic (stick together)
- Mass of object 1 (m₁): 2 kg
- Initial velocity of object 1 (u₁): 3 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -2 m/s
- Show velocities in: Kilometers per hour (km/h)
Find: Common final velocity
Answer
10.
Problem 10.
- Type of collision: Perfectly inelastic (stick together)
- Mass of object 1 (m₁): 2 kg
- Initial velocity of object 1 (u₁): 2 m/s
- Mass of object 2 (m₂): 1 kg
- Initial velocity of object 2 (u₂): -1 m/s
- Show velocities in: Miles per hour (mph)
Find: Common final velocity
Answer

Answer key: Elastic Collision worksheet

1. Common final velocity: 2 m/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $3 \times 3 + 1 \times (−1)$ $= 8 kg·m/s$
2. The objects stick and move as one, so share the momentum over their combined mass
  $\frac{8}{3 + 1}$ $= 2 m/s$
3. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 3 \times 3^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 14 J$
4. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 3 \times 2^{2} + 0.5 \times 1 \times 2^{2}$ $= 8 J$
2. Common final velocity: -1.09 ft/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $1 \times 3 + 2 \times (−2)$ $= −1 kg·m/s$
2. The objects stick and move as one, so share the momentum over their combined mass
  $\frac{−1}{1 + 2}$ $= −0.333333 m/s$
3. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 1 \times 3^{2} + 0.5 \times 2 \times {(−2)}^{2}$ $= 8.5 J$
4. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 1 \times {(−0.333333)}^{2} + 0.5 \times 2 \times {(−0.333333)}^{2}$ $= 0.166667 J$
5. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{−0.333333}{0.3048}$ $= −1.093613 ft/s$
3. Final velocity of object 1 (v₁): -1.09 ft/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $2 \times 3 + 1 \times (−2)$ $= 4 kg·m/s$
2. No energy is lost, so a standard formula gives object 1's velocity after the bounce (v₁)
  $\frac{(2 - 1) \times 3 + 2 \times 1 \times (−2)}{2 + 1}$ $= −0.333333 m/s$
3. The matching formula gives object 2's velocity after the bounce (v₂)
  $\frac{(1 - 2) \times (−2) + 2 \times 2 \times 3}{2 + 1}$ $= 4.666667 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 2 \times 3^{2} + 0.5 \times 1 \times {(−2)}^{2}$ $= 11 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 2 \times {(−0.333333)}^{2} + 0.5 \times 1 \times {4.666667}^{2}$ $= 11 J$
6. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{−0.333333}{0.3048}$ $= −1.093613 ft/s$
7. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{4.666667}{0.3048}$ $= 15.310586 ft/s$
4. Final velocity of object 1 (v₁): 0 mph
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $3 \times 4 + 2 \times (−1)$ $= 10 kg·m/s$
2. No energy is lost, so a standard formula gives object 1's velocity after the bounce (v₁)
  $\frac{(3 - 2) \times 4 + 2 \times 2 \times (−1)}{3 + 2}$ $= 0 m/s$
3. The matching formula gives object 2's velocity after the bounce (v₂)
  $\frac{(2 - 3) \times (−1) + 2 \times 3 \times 4}{3 + 2}$ $= 5 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 3 \times 4^{2} + 0.5 \times 2 \times {(−1)}^{2}$ $= 25 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 3 \times 0^{2} + 0.5 \times 2 \times 5^{2}$ $= 25 J$
6. Change the answer to mph, the unit you picked: divide by 0.44704
  $\frac{0}{0.44704}$ $= 0 mph$
7. Change the answer to mph, the unit you picked: divide by 0.44704
  $\frac{5}{0.44704}$ $= 11.184681 mph$
5. Final velocity of object 1 (v₁): 1.13 m/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $2 \times 3 + 1 \times (−1)$ $= 5 kg·m/s$
2. Some energy is lost, so use the bounce factor e to find object 1's velocity after (v₁)
  $\frac{5 + 1 \times 0.4 \times (−1 - 3)}{2 + 1}$ $= 1.133333 m/s$
3. Do the same for object 2's velocity after (v₂)
  $\frac{5 + 2 \times 0.4 \times (3 - (−1))}{2 + 1}$ $= 2.733333 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 2 \times 3^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 9.5 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 2 \times {1.133333}^{2} + 0.5 \times 1 \times {2.733333}^{2}$ $= 5.02 J$
6. Final velocity of object 1 (v₁): -1 m/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $1 \times 3 + 1 \times (−1)$ $= 2 kg·m/s$
2. No energy is lost, so a standard formula gives object 1's velocity after the bounce (v₁)
  $\frac{(1 - 1) \times 3 + 2 \times 1 \times (−1)}{1 + 1}$ $= −1 m/s$
3. The matching formula gives object 2's velocity after the bounce (v₂)
  $\frac{(1 - 1) \times (−1) + 2 \times 1 \times 3}{1 + 1}$ $= 3 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 1 \times 3^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 5 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 1 \times {(−1)}^{2} + 0.5 \times 1 \times 3^{2}$ $= 5 J$
7. Final velocity of object 1 (v₁): 4.92 ft/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $3 \times 4 + 1 \times (−1)$ $= 11 kg·m/s$
2. No energy is lost, so a standard formula gives object 1's velocity after the bounce (v₁)
  $\frac{(3 - 1) \times 4 + 2 \times 1 \times (−1)}{3 + 1}$ $= 1.5 m/s$
3. The matching formula gives object 2's velocity after the bounce (v₂)
  $\frac{(1 - 3) \times (−1) + 2 \times 3 \times 4}{3 + 1}$ $= 6.5 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 3 \times 4^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 24.5 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 3 \times {1.5}^{2} + 0.5 \times 1 \times {6.5}^{2}$ $= 24.5 J$
6. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{1.5}{0.3048}$ $= 4.92126 ft/s$
7. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{6.5}{0.3048}$ $= 21.325459 ft/s$
8. Final velocity of object 1 (v₁): 1.64 ft/s
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $2 \times 2 + 1 \times (−1)$ $= 3 kg·m/s$
2. Some energy is lost, so use the bounce factor e to find object 1's velocity after (v₁)
  $\frac{3 + 1 \times 0.5 \times (−1 - 2)}{2 + 1}$ $= 0.5 m/s$
3. Do the same for object 2's velocity after (v₂)
  $\frac{3 + 2 \times 0.5 \times (2 - (−1))}{2 + 1}$ $= 2 m/s$
4. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 2 \times 2^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 4.5 J$
5. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 2 \times {0.5}^{2} + 0.5 \times 1 \times 2^{2}$ $= 2.25 J$
6. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{0.5}{0.3048}$ $= 1.64042 ft/s$
7. Change the answer to ft/s, the unit you picked: divide by 0.3048
  $\frac{2}{0.3048}$ $= 6.56168 ft/s$
9. Common final velocity: 4.8 km/h
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $2 \times 3 + 1 \times (−2)$ $= 4 kg·m/s$
2. The objects stick and move as one, so share the momentum over their combined mass
  $\frac{4}{2 + 1}$ $= 1.333333 m/s$
3. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 2 \times 3^{2} + 0.5 \times 1 \times {(−2)}^{2}$ $= 11 J$
4. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 2 \times {1.333333}^{2} + 0.5 \times 1 \times {1.333333}^{2}$ $= 2.666667 J$
5. Change the answer to km/h, the unit you picked: divide by 0.277777777778
  $\frac{1.333333}{0.277777777778}$ $= 4.8 km/h$
10. Common final velocity: 2.24 mph
1. Multiply each mass by its velocity (speed, with a minus sign for leftward) and add. This total momentum p stays the same after the crash
  $2 \times 2 + 1 \times (−1)$ $= 3 kg·m/s$
2. The objects stick and move as one, so share the momentum over their combined mass
  $\frac{3}{2 + 1}$ $= 1 m/s$
3. Find the kinetic energy (energy of motion) before: half of each mass times its speed squared, added up
  $0.5 \times 2 \times 2^{2} + 0.5 \times 1 \times {(−1)}^{2}$ $= 4.5 J$
4. Do the same with the final speeds, to see how much energy is left
  $0.5 \times 2 \times 1^{2} + 0.5 \times 1 \times 1^{2}$ $= 1.5 J$
5. Change the answer to mph, the unit you picked: divide by 0.44704
  $\frac{1}{0.44704}$ $= 2.236936 mph$