Momentum, Collisions and Explosions Legacy Problem #2 Guided Solution
Problem*
A bicycle has a momentum of \(24 \unit{\meter\per\second}\). What momentum would the bicycle have if it had …
- Twice the mass and was moving at the same speed?
- The same mass and was moving with twice the speed?
- One-half the mass and was moving with twice the speed?
- The same mass and was moving with one-half the speed?
- Three times the mass and was moving with one-half the speed?
- Three times the mass and was moving with twice the speed?
Audio Guided Solution
The equation for momentum is that momentum is equal to the mass of an object times the velocity of an object, or p, for momentum, is equal to m times v. The fact that this is the equation for momentum tells us that momentum depends upon two quantities, mass and velocity. Not only that, momentum depends directly upon mass and directly upon velocity. Any alteration that is made in the mass of an object would produce the same change in the momentum of an object. And similarly, any alteration which is made in the velocity or speed of an object would likewise affect the momentum by the same factor. So in this problem, we're told a bicycle has 24 units of momentum. And then, what would be the momentum of a bicycle with twice the mass? Well, if it had twice the mass and was still moving at the same speed, then momentum depends directly upon mass. Twice the mass means twice the momentum, and thus 48 kilograms times meter per second. If instead of having twice the mass, it had the same mass but was moving with twice the speed, we'd have the same effect, since momentum is dependent directly upon the speed of an object. So we would still have a two-fold increase in the original momentum, and we'd have 48 kilograms times a meter per second. In part c, we're told an object's moving with one-half the mass, which in itself would cause the momentum to decrease to 12. But it's moving with twice the speed, which would therefore offset the one-half mass factor, and we would end up having the same momentum, since the two factors offset one another. In part d, we're asked to find the momentum of the bicycle if it had the same mass but was moving with one-half the speed. In this problem, we have to consider the fact that mass isn't altered, but the speed is. In fact, it's decreased, and it's decreased by a factor of two. As such, the momentum would decrease by a factor of two, and instead of having 24 units, we would have 12 units of momentum. In part e, we're told we have three times the mass and one-half the speed. The fact that the mass is three times as much means we have to take the 24 and multiply it by 3. But that's not all, because we also have to account for the fact that the speed is one-half of what it originally was. So one-half the speed would mean that the original 24 would also have to be divided by 2. So you have two effects here, one which causes a three-fold increase, and the other which causes a two-fold decrease. You end up taking 24 and multiplying by 3 and dividing by 2 to get the 36. And finally, in f, we have to take the 24 units of momentum, and we have to multiply by 3 for the fact that the mass is triple. And then we have to multiply by 2 once more for the fact that the speed has increased. That's 24 times 3 times 2, which comes out to be 144.
Solution
- \(48\unit{\kg\meter\per\second}\)
- \(48\unit{\kg\meter\per\second}\)
- \(24\unit{\kg\meter\per\second}\)
- \(12\unit{\kg\meter\per\second}\)
- \(36\unit{\kg\meter\per\second}\)
- \(144\unit{\kg\meter\per\second}\)
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!
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