Momentum, Collisions and Explosions Legacy Problem #17 Guided Solution
Problem*
To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broom in an effort to scare the crow away, it pushes off the 670-gram feeder with a takeoff speed of 1.5 m/s. Determine the speed at which the feeder initially recoils backwards.
Audio Guided Solution
A good problem solver reads the problem carefully and develops a mental picture of what is happening, identifies the known quantities and the unknown quantities, and then plots out a strategy using conceptual reasoning skills to get from the known to the unknown information. The picture we have here is of a black crow who is at rest on a bird feeder. When scared by Mr. H running out with a broom, the black crow pushes off the bird feeder. With a take-off speed of 1.5 meters per second, we're given the mass of the black crow and the mass of the bird feeder. We're asked to determine the speed of the feeder after the crow pushes off. The concept we'll use to get from the known to the unknown information is that in any given explosion or collision, the momentum change of one object is equal to the momentum change of the other object. So here, the momentum change of the black crow should be equal to the momentum change of the feeder. One of them will gain forward momentum, the other will gain backwards momentum. So the way we'll approach it is we'll begin by finding the momentum change of the crow. After all, we know the mass, and we know its pre- and post-impulse speed. Before pushing off the feeder, it was at rest, 0 meters per second, and afterwards, its speed was 1.5 meters per second. This is a velocity change of 1.5 meters per second. So if we want to find the momentum change of the black crow, we just go mass times this velocity change, 450 grams times 1.5 meters per second. This gives us a value of 675 grams times meter per second as the momentum change of the black crow. The momentum change of the feeder would be the same magnitude, only in the opposite direction. So we can say that the momentum change of the feeder, 670 grams times delta V, is equal to the same momentum change of 675 grams times meter per second. When you divide through by 670, you get 1.0075 as the velocity change of the feeder. We can round this to two digits, and it becomes 1.0 meters per second.
Solution
1.0 m/s
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(m = 1.50 \unit{kg}\), \(v_i = 2.68 \unit{\meter\per\second}\), \(F = 4.98 \unit{\newton}\), \(t = 0.133 \unit{\second}\), \(v_f = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
Get more information on the topic of Momentum, Collisions and Explosions at The Physics Classroom Tutorial.