Momentum, Collisions and Explosions Legacy Problem #6 Guided Solution
Problem*
An interesting story is often told of baseball star Johnny Bench when he was a rookie catcher in 1968. During a Spring Training game, he kept signaling to star pitcher Jim Maloney to throw a curve ball. Maloney continuously shook off Bench's signal, opting to throw fastballs instead. The rookie catcher walked to the mound and told the veteran Maloney that his fastball wasn't fast enough and that he should throw some curve balls. Bench again signaled for a curve. Maloney shook of the signal and threw a fastball. Before the ball reached the plate, Bench took off his glove; he then caught the pitch barehanded.
- Determine the impulse required to stop a \(0.145\unit{kg}\) baseball moving at \(35.7\unit{\meter\per\second}\) (\(\units{80.0}{\unitfrac{\text{mile}}{\text{hour}}}\)).
- If this impulse is delivered to the ball in \(0.020\unit{s}\), then what is magnitude of the force acting between the bare hand and the ball?
Audio Guided Solution
In this problem, there is a collision between a baseball player's hand and a high-speed baseball. The mass of the baseball is given, 0.145 kilograms, and the speed of the baseball before the collision is also given, 35.7 meters per second. The baseball is brought to a stop and we are asked to determine the impulse. Now if you understand impulse to be force times time, then you may be looking for a long time to find the force and the time. Now for certain the impulse is the force multiplied by the time, but the impulse also is equal to the momentum change. And so if we know the momentum change, we automatically know the impulse. And here the momentum change can be calculated if we take the mass of the baseball multiplied by the velocity change. The baseball changes its velocity from an original speed of 35.7 meters per second to a zero meters per second. So the velocity change is negative 35.7 meters per second. Taking that value for velocity change and multiplying by 0.145 gives us a value for the momentum change, and thus for the impulse. That value is negative 5.1765 newtons times a second, or equivalently kilograms times a meter per second. Now in the second part of the problem, we have to determine the force acting on the ball if this impulse is delivered to it in 0.020 seconds. That is, if in two one hundredths of a second the ball comes to a stop, determine what the force is. In order to figure this out, we have to use the definition of impulse as being equal to the force times the time. We're given the time here as 0.020 seconds, and we've just calculated the impulse to be 5.1765 newtons times a second. So if I set that number equal to F times 0.02, I can solve for F. And when I do, I get a value of about 258.825 newtons. I can round that to two significant digits consistent with the 0.020 seconds that's given in this problem. So that would be 260 newtons.
Solution
- \(-5.18 \unit{\newton\second}\)
- \(260\unit{\newton}\) (rounded from \(259\unit{\newton}\))
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.
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