Consider a scenario during which a 2.00kg mass (A) moving at 2.00 m/s strikes an

Question

Consider a scenario during which a 2.00kg mass (A) moving at 2.00 m/s strikes another 2.00 is mass (B) which is at rest. Both are confined to move along the same straight line. Three possible outcomes of the collision that conserve momentum are: A) they stick together and move off at 1.00 m/s B) they bounce off of one another and move off in the same direction, one (A) at 0.50 m/s and the other (B) at 1.50 m/s C) (A) comes to rest and (B) moves off at 2.00 m/s
In each case the momentum before the collision is: (2.00 kg) (2.00 m/s) = 4.00 kg * m/s
1. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision. 2. In each of the three cases, find the kinetic energy lost and characterize the collision as elastic, partially inelastic, or totally inelastic. The kinetic energy before the collision is (1/2)(2.00 kg)(2.00 m/s)^2 = 4.00 kg * m^2/s^2 = 4.00 J.
PHYSICS 1101 EXPERIMENT #7 MOMENTUM AND COLLISIONS PREPARATION SHEET Lab Assistant Name Date Submitted Lab Day & Hour for the experiment by doing the tasks on this sheet and studying the instructions for the Prepare experiment. TURN IN THIS SHEETAT THE BEGINNING OF THE LABORATORY PERIOD Study the material on collisions, kinetic energy, momentum, and momentum conservation in your text. Pay particular attention to totally inelastie collisions. Study the instructions for this experiment. Work the following problems. They are similar to those you will encounter in the experiment Consider a scenario during which a 2.00 kg mass (A) moving at 2.00 m/s strikes another 2.00 kg mass (B) which is at rest. Both are confined to move along the same straight line. Three possible outcomes of the collision that conserve momentum are: a) they stick together and move off at 1.00 m/s. b) they bounce off one another and move off in the same direction, one (A) at 0.50 m/s and the other (B) at 1.50 m/s. c) (A) comes to rest and (B) moves off at 2.00 m/s. In each case the momentum before the collision is: (2.00 kg)(2.00 m/s)-4.00 kg-m/s. 1. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision Case a) P (after)- Case b) P (after)- Case c) P (after) Revised 12-19-08 Experiment 7-Page 1 All Rights Reserved

Explanation / Answer

Part 1:

a)

p_after = (2kg + 2kg) * 1m/s

p_after = 4 kg-m/s

b)

p_after = (2kg*0.5m/s) + (2kg*1.5m/s)

p_after = 4 kg-m/s

c)

p_after = 0 + 2kg*2m/s

p_after = 4 kg-m/s

Part 2:

a)

KE_after = ½*4kg*(1m/s)²

KE_after = 2 J

change = 2 J lost (completely inelastic)

b)

KE_after = ½*2kg*((0.5m/s)² + (1.5m/s)²)

KE_after = 2.5 J

Change = 1.5 J lost (partially elastic)

c)

KE_after = ½*2kg*(2m/s)²

KE_after = 4 J

Change = 0 J lost (completely elastic)

Part 3:

KE_after = 4 J

p_after = 2kg * 1.414m/s + 2kg * 1.414m/s

p_after = 5.656 kg-m/s

impossible, the conservation of energy impossible because the momenta are not conserved. Momentum has been created

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---