Momentum, Collisions and Explosions Legacy Problem #7 Guided Solution
Problem*
While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds.
- Determine the force acting upon Logan's body.
- If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds. Determine what the force on his body would have been for such an abrupt collision.
Audio Guided Solution
Here in this problem, Logan is having his momentum changed from an original value to zero in 0.16 seconds. We're asked to determine the force which is acting upon Logan's body. What we know is the original velocity before the collision was 7.6 and that the body came to rest and that the mass of the body is 74 kilograms and the time it took was 0.16 seconds. So we have an equation in this unit known as the impulse momentum change equation that states that F times the T is equal to the M times the delta V. So we can take that equation and we can divide through both sides of it by T which gives us a new equation, T equal M times delta V all divided by the T. In this question, the M is 74 kilograms and the delta V is negative 7.6. We can take those two numbers and multiply and then divide by 0.16 seconds and what we get is 3,515 newtons. We can round that to two significant digits, 3.5 times 10 to the third newtons. Now in part B, we're asked what happens if instead of hitting the padded wall, Logan happened to hit the concrete wall in the gymnasium. Moving at the same speed and having the same original mass, what would be the new force acting upon his body in a more abrupt collision given that the time that it took to stop would be 0.0080 seconds instead of the 0.16. Now you might notice right away that that new time is one twentieth the previous time. We're going to do the problem the same way we did the previous one. We're going to go 74 times negative 7.6 and then divide through by 0.008. Divide through by a number which is actually one twentieth the original number. When we do that, we get a number of 70,300 newtons or rounded to two digits it becomes 7.0 times 10 to the fourth newtons. You might notice that this new number that we've calculated for force is 20 times the original value. The message here is if you can make the time of a collision last 20 times as long, then you'll make the force on the object one twentieth of the original value. And that's physics for a better living.
Solution
- 3.5 x 103 N
- 7.0 x 104 N
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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