REVIEW LAST DAY WITH TWO ANIMATIONS:
The Cart and The Brick
Part A
The Cart and The Brick
Part B
The Law of Momentum Conservation
Using Equations as a Recipe for Algebraic Problem-Solving
Total system momentum is conserved for collisions between objects in an isolated system.
The momentum lost by one object is equal to the momentum gained by another object.
This law becomes a powerful tool in physics because it allows for predictions of the before- and after-collision velocities (or mass) of an object.
EXAMPLE: A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.
ANSWER: Such a motion can be considered as a collision between a person and a medicine ball.
Before the collision, the ball has momentum and the person does not.
The collision causes the ball to lose momentum and the person to gain momentum.
After the collision, the ball and the person travel with the same velocity (v) across the ice.
If it can be assumed that the affect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system.
Momentum should be conserved and the post-collision velocity (v) can be determined using a momentum table as shown below.
Before Collision
/After Collision
Person
/ 0 / (60 kg) • vMedicine ball
/ (15 kg) • (20 km/hr)= 300 kg • km/hr / (15 kg) • v
Total
/ 300 kg • km/hr / 300Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision.
The expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball.
To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:
60 • v + 15 • v = 300
75 • v = 300
v = 4 km/hr
. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision. (NOTE: The unit km/hr is the unit on the answer since the original velocity as stated in the question had units of km/hr.)
EXAMPLE: A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum.
If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.
ANSWER: Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum.
After the collision, the ball and the mitt move with the same velocity (v) .
The collision between the ball and the catcher's mitt occurs in an isolated system, total system momentum is conserved.
Cont’ next page
Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt).
Before Collision
/After Collision
Ball
/ 0.15 kg • 45 m/s = 6.75 kg•m/s / (0.15 kg) • vCatcher's Mitt
/ 0 / (0.25 kg) • vTotal
/ 6.75 kg•m/s / 6.75 kg•m/sSince momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision.
0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s
0.40 kg • v = 6.75 kg•m/s
v = 16.9 m/s
Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.
The two collisions above are examples of inelastic collisions.
Technically, an inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. In an inelastic collision, the kinetic energy of the colliding objects is transformed into other non-mechanical forms of energy such as heat energy and sound energy.
The subject of energy will be treated in a later unit of The Physics Classroom.
To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision.
Now we will consider the analysis of a collision in which the two objects do notstick together. In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together.
EXAMPLE: A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.
ANSWER: In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).
The collision can be analyzed using a momentum table similar to the above situations.
Before Collision
/After Collision
Truck
/ 3000 • 10 = 30 000 / 3000 • vCar
/ 0 / 1000 • 15 = 15 000Total
/ 30 000 / 30 0003000*v + 15 000 = 30 000
3000*v = 15 000
v = 5.0 m/s
The truck's velocity immediately after the collision is 5.0 m/s. As predicted, the truck has lost momentum (slowed down) and the car has gained momentum.
TRUCK CAR COLLISIONS (6 ANIMATIONS) at
Using Equations as a Guide to Thinking
The questions which follow provide a real test of your conceptual understanding of momentum conservation in collisions.
EXAMPLE: Suppose that you have joined NASA and are enjoying your first space walk. You are outside the space shuttle when your fellow astronaut of approximately equal mass is moving towards you at 2 m/s (with respect to the shuttle). If she collides with you and holds onto you, then how fast (with respect to the shuttle do you both move after the collision?
This problem could be solved in the usual manner with a momentum table; the variable m could be used for the mass of the astronauts or any random number could be used for the mass of the astronauts (provided each astronaut had the same mass).
In the process of solving the problem, the mass would cancel out of the momentum conservation equation and the post-collision velocities could be determined.
However, there is a more conceptual means of solving this problem. In order for the momentum before the collision to be equal to the momentum after the collision, the after collision velocity must be smaller than the before collision velocity.
How many times smaller must it be?
By what factor must the velocity be decreased?
Before the collision, the amount of mass in motion is m; after the collision, the amount of mass in motion is 2•m. The amount of mass in motion has doubled as the result of the collision.
If the mass is increased by a factor of two, then the velocity must be decreased by a factor of 2. The before-collision velocity was 2 m/s so the after-collision velocity must be one-half this value: 1 m/s. Each astronaut is moving with a velocity of 1 m/s after the collision.
The equation for momentum (p=m*v) becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity.
Of course, it is instructive to point out that this form of problem-solving is limited to situations in which one of the two objects is at rest before the collision and both objects move at the same speed after the collision.
ASTRONAUT CATCH ANIMATION at
EXAMPLE: A large fish is in motion at 2 m/s when it encounters a smaller fish which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish (and the smaller fish) after the collision?
ANSWER: Use your mind, not math to figure this problem out. The amount of mass in motion is increased from 3m to 4m (3m + m); that is mass is increased by a factor of 4/3 (1.33). To conserve momentum, an increase in mass by a factor of 1.33 must be accompanied by a decrease in velocity by a factor of 1.33. Thus divide the original velocity of 2 m/s by 1.33.
Mathematically:
3*m*2 = (3*m + m)*v
6*m = 4*m*v
(6*m) / (4*m) = v
v = 1.5 m/s
THE FISH CATCH (2 ANIMATIONS) at
Inelastic Collision
EXAMPLE: A railroad diesel engine has five times the mass of a boxcar. A diesel coasts backwards along the track at 4 m/s and couples together with the boxcar (initially at rest). How fast do the two trains cars coast after they have coupled together?
Answer: Use your mind not equations: The amount of mass in motion is increased from 5m to 6m (5m+m). That is, the total mass which is moving is increased by a factor of 6/5 (or 1.20). To conserve momentum, an increase in mass by a factor of 1.20 must be accompanied by a decrease in velocity by a factor of 1.20. Thus, divide the original velocity of 4 m/s by 1.2.
Still another method:
5*m*4 = (5*m + m)*v
20*m = 6*m*v
(20*m) / (6*m) = v
3.3 m/s = v
The Diesel and Flat Car at
Inelastic Collision
MOMENTUM LESSON 4 HOMEWORK
Do p. 23 Conceptual Physics Worksheet
1. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 92.8 cm/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 21.6 cm/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.
2. A 25.0-gram bullet enters a 2.35-kg watermelon and embeds itself in the melon. The melon is immediately set into motion with a speed of 3.82 m/s. The bullet remains lodged inside the melon. What was the entry speed of the bullet? (CAUTION: Be careful of the units on mass.)
3. A 25.0-gram bullet enters a 2.35-kg watermelon with a speed of 217 m/s and exits the opposite side with a speed of 109 m/s. If the melon was originally at rest, then what speed will it have as the bullet leaves its opposite side? (CAUTION: Be careful of the units on mass.)
4. In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (Cart B) which is initially at rest. The 0.500-kg cart rebounds with a speed of 45 cm/s in the opposite direction. Determine the post-collision speed of the 1.50-kg cart.
5. A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign on the velocity of the two vehicles.)
6. During a goal-line stand, a 75-kg fullback moving eastward with a speed of 8 m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the post-collision velocity of the two players. (CAREFUL: Be cautious of the +/- sign on the velocity of the two players.)
7. An astronaut at rest in space fires a thruster pistol that expels 35 g of hot gas at 875 m/s. The combined mass of the astronaut and pitol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol?
8. A 12.0 g rubber bullet travels at a velocity of 150 m/s, hits a stationary 8.5 kg concrete block resting on a frictionless surface, and ricochets in the opposite direction with a velocity of -1.0 x 102 m/s. How fast will the concrete block be moving?
9. A 50.0 g projectile is launched with a horizontal velocity of 647 m/s from a 4.65 kg launcher moving in the same direction at 2.00 m/s. What is the launcher’s velocity after the launch and in what direction?
10. Kofi, with mass 42.00 kg, is riding a skateboard with a mass of 2.00 kg and traveling at 1.20 m/s. Kofi jumps off and the skateboard stops dead in its tracks. In what direction and with what velocity did he jump?
11. A 75 kg fullback, running at 5.0 m/s, attempts to dive directly across the goal line for a touchdown. Just as he reaches the line, he is met head on in midair by two 75 kg linebackers, both moving at 2.5 m/s in the direction opposite the fullback. They all become entagled as one mass. Solve this problem with thinking, not mathematical calculations.
(a)Sketch the event, identifying the “before” and “after” situations.
(b)What is the velocity of the football players after the collision?
(c)Does the fullback score a touchdown?
HOMEWORK KEY
1.7.00 cm/s
2.363 m/s
3.1.15 m/s
4.58.0 cm/s
5.1.25 km/h, right
6.1.14 m/s, East
7.0.36 m/s, opposite direction as hot gas
8.1.6 m/s
9.-4.94 m/s or 4.94 m/s in opposite direction to projectile
10.1.26 m/s in direction of original motion.
11.zero, no
MOMENTUM LESSON 4 HOMEWORK
Answers
1. The problem can be solved using a momentum table:
2. The problem can be solved using a momentum table:
3. The problem can be solved using a momentum table:
4. The problem can be solved using a momentum table:
5. The problem can be solved using a momentum table:
6. The problem can be solved using a momentum table:
7. Before the pistol is fired, all parts of the system are at rest; thus, the initial momentum is zero. After the pistol is fired, the momentum of the astronaut is equal in magnitude, but opposite in direction to the momentum of the gas leaving the pistol.
Before Collision / After CollisionAstronaut / 0 / 0.035*875
Gas / 0 / 84*v
Total / 0 / 0
0 = 0.035(-875) + 84*v
v = 0.36 m/s, opposite direction to hot gas
8.
Before Collision
/After Collision
Bullet
/0.012(150)
/0.012(-1.0 x 102)
Concrete Block
/0
/8.5(v)
Total
/0.012(150) = 1.8
/1.8
1.8 = 0.12(-1.0 x 102) + 8.5v
v = 1.6 m/s
9.
Before Collision
/After Collision
Projectile
/0.050(2.00)
/0.050(647)
Launcher
/4.65(2.00)
/4.65(v)
Total
/0.1 + 9.3 = 9.4
/9.4
9.4 = 0.050(647) + 4.65v
9.4 – 32.35 = 4.65v
v = -4.94 m/s
10.
Before Collision
/After Collision
Kofi
/42.00(1.20)
/42.00(v)
Skateboard
/2.00(1.20)
/0
Total
/50.4 + 2.40 = 52.8
/52.8
52.8 = 42.00(v) + 0
v = 1.26 m/s in direction of original motion. As Kofi pushes off the skateboard, he pushes back on the skateboard and this force stops the momentum of the skateboard.
11. Before the collision, the amount of mass in motion is 75 kg at 5.0 m/s in one direction, and 150 kg at 2.5 m/s in the opposite direction. The combined momentum of these two masses = 0 since mass is doubled, but velocity is halved for the opposite direction.
Therefore the momentum of the group of football players after collision also has to be zero, therefore the velocity of the football players is zero and the fullback does not score a touchdown.