Question

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 7.83 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.1 m/s along the same line. When they meet, they grab each other and hang on.

a. What is their velocity immediately thereafter?
b. What fraction of their original kinetic energy is mechanical energy after their collision?
c. That was so much fun that the boys repeat their collision with the same original velocities, this time moving along parallel lines 1.20 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Find their angular speed.
d. What fraction of their original kinetic energy is still mechanical energy after they link arms?
e. Why are the answers to parts (b) and (d) so different?

Answer 1

Answer:

a) Their velocity is 0.516 m/s

b) The fraction of their original kinetic energy is 0.0032

c) Their angular speed is 7.54 rad/s

d) The fraction of their kinetic energy is 0.263

e) The responses are different because head-on collisions between two similarly sized objects are inefficient. If the children were the same size and each had the same speed, the conservation of the moment would say that they would lose their kinetic energy.

Explanation:

a) The final speed of the boys using the expression of conservation of momentum is equal to:

$P_{1} =P_{2} \\(45*7.83)+(31*(-10.1))=(45+31)v_{2} \\v_{2} =\frac{(45*7.83)-(31*10.1)}{45+31} =0.516m/s$

b) The fraction of energy that remains is equal to:

$\frac{E_{k,final}}{E_{k,initial} } =\frac{\frac{1}{2}*0.516^{2} *(45+31) }{(\frac{1}{2}*45*7.83^{2})+(\frac{1}{2}*35*10.1^{2} )} =0.0032$

c) The angular momentum is:

$L=mvr=\frac{4*7.83*1.2}{2} +\frac{31*10.1*1.2}{2} =206.65kgm^{2} /s$

The moment of inertia is:

$I=mr^{2} =(45*0.6^{2} )+(31*0.6^{2} )=27.4kgm^{2}$

The angular velocity is equal to:

$w=\frac{L}{I} =\frac{206.65}{27.4} =7.54rad/s$

d) The fraction of kinetic energy is:

$\frac{E_{k,final linear}+E_{k,final angular}}{E_{k,initial} } =\frac{\frac{1}{2}*0.25^{2} *(45+31)+\frac{1}{2}*27.4 *7.54^{2} }{(\frac{1}{2}*45*7.83^{2})+(\frac{1}{2}*35*10.1^{2} )} =0.263$

e) The responses are different because head-on collisions between two similarly sized objects are inefficient. If the children were the same size and each had the same speed, the conservation of the moment would say that they would lose their kinetic energy.

Answer 2

The velocity of the boys immediately thereafter will be 0.516m/s.

How to calculate the velocity?

The velocity will be calculated thus based on the information given:

(45 × 7.83) + (31 × -10.1) = (45 + 31)V

V = (45 × 7.83) - (31 × 10.1) / (45 + 31)

V = 0.516m/s

The fraction of their original kinetic energy is mechanical energy after their collision will be 0.0032.

The angular momentum will be:

= (4 × 7 × 0.83 × 1.2)/2 + (31 × 10.1 × 1.2)/2

= 206.65 kgm/s.

The moment of inertia will be:

I = mr²

I = (45 × 0.6²) + (31 × 0.6²)

I = 27.4 kgm²

The angular velocity will be calculated thus:

w = 206.65/27.4 = 7.54 rad/s

In conclusion, the angular velocity is 7.54 rad/s.

Learn more about velocity on:

https://brainly.com/question/25749514