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Chapter 7: Problem 15

A boy of mass \(60.0 \mathrm{kg}\) is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was \(8.0 \mathrm{m}\) above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of \(0.40 \mathrm{s}\). (a) What is his change in momentum during the 0.40 -s interval? (b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.] (c) What is the average force on the net due to the boy during the interval?

Short Answer

Expert verified
Question: Calculate the initial velocity of the boy just before he lands on the net, his change in momentum, the impulse on the net, and the average force on the net during his deceleration process. Answer: Using the provided information and the steps outlined above, we find that the initial velocity of the boy is approximately \(12.6 \mathrm{m/s}\). The change in momentum of the boy is approximately \(-756 \mathrm{kg\cdot m/s}\), which is the same as the impulse on the net due to the boy. The average force on the net due to the boy during the deceleration process is approximately \(-1890 \mathrm{N}\).

Step by step solution

01

Calculate the initial velocity of the boy

First, we need to find the initial velocity of the boy just before he lands on the net. Since the boy is in free fall, we can use the kinematic equation to find his initial velocity: \(v^2 = u^2 + 2as\) where \(v\) is the final velocity, \(u\) is the initial velocity (which is \(0\) as he starts from rest), \(a\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), and \(s\) is the vertical displacement (\(8.0 \mathrm{m}\)). After rearranging the equation, we get: \(v = \sqrt{2as}\)
02

Calculate the final velocity of the boy

Now, we need to find the final velocity of the boy after he comes to a complete stop. Since he stops in the net, his final velocity is \(0 \mathrm{m/s}\).
03

Calculate the change in momentum of the boy

To find the change in momentum of the boy, we can use the formula: \(\Delta p = m(v_f - v_i)\) where \(m\) is the mass of the boy (\(60.0 \mathrm{kg}\)), \(v_f\) is the final velocity (\(0 \mathrm{m/s}\)), and \(v_i\) is the initial velocity calculated in step 1. Plug the values into the formula to find the change in momentum.
04

Calculate the impulse on the net due to the boy

The impulse on the net is equal to the change in momentum of the boy. Since we found the change in momentum in step 3, we can use the same value for the impulse. This is because the impulse applied by the net should be equal and opposite to the momentum change of the boy.
05

Calculate the average force on the net due to the boy

To find the average force on the net during the interval, we can use the impulse-momentum theorem, which states that: \(I = Ft\) where \(I\) is the impulse, \(F\) is the average force, and \(t\) is the time interval (\(0.40 \mathrm{s}\)). Rearranging the formula, we get: \(F = \frac{I}{t}\) We can use the impulse calculated in step 4 and divide it by the given time interval to find the average force on the net due to the boy.

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Most popular questions from this chapter

In a circus trapeze act, two acrobats actually fly through the air and grab on to one another, then together grab a swinging bar. One acrobat, with a mass of \(60 \mathrm{kg},\) is moving at \(3.0 \mathrm{m} / \mathrm{s}\) at an angle of \(10^{\circ}\) above the horizontal and the other, with a mass of $80 \mathrm{kg},\( is approaching her with a speed of \)2.0 \mathrm{m} / \mathrm{s}$ at an angle of \(20^{\circ}\) above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive \(x\) -axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive \(x\)- and \(y\)-directions.
A sled of mass \(5.0 \mathrm{kg}\) is coasting along on a frictionless ice- covered lake at a constant speed of \(1.0 \mathrm{m} / \mathrm{s} .\) A \(1.0-\mathrm{kg}\) book is dropped vertically onto the sled. At what speed does the sled move once the book is on it?
A projectile of 1.0-kg mass approaches a stationary body of \(5.0 \mathrm{kg}\) at \(10.0 \mathrm{m} / \mathrm{s}\) and, after colliding, rebounds in the reverse direction along the same line with a speed of $5.0 \mathrm{m} / \mathrm{s} .$ What is the speed of the 5.0 -kg body after the collision?
Verify that the SI unit of impulse is the same as the SI unit of momentum.

Jane is sitting on a chair with her lower leg at a \(30.0^{\circ}\) angle with respect to the vertical, as shown. You need to develop a computer model of her leg to assist in some medical research. If you assume that her leg can be modeled as two uniform cylinders, one with mass \(M=\) \(20 \mathrm{kg}\) and length \(L=35 \mathrm{cm}\) and one with mass \(\quad m=\) \(10 \mathrm{kg}\) and length \(l=40 \mathrm{cm},\) where is the CM of her leg?
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