Question

In: Physics

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45...

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 7.95 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.9 m/s along the same line. When they meet, they grab each other and hang on. (a) What is their velocity immediately thereafter? magnitude m/s direction (b) What fraction of their original kinetic energy is still mechanical energy after their collision? % (c) That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.02 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch. Find the velocity of their center of mass. magnitude m/s direction (d) Find their angular speed. rad/s (e) What fraction of their original kinetic energy is still mechanical energy after they link arms? % (f) Why are the answers to parts (b) and (e) so different? This answer has not been graded yet.

Solutions

Expert Solution

A)

In a head-on collision where the bodies stick together kinetic energy is not conserved but momentum is We can calculate the final speed of the boys using conservation of momentum.

P1 = P2

45*7.95 - 31*10.9 = ( 45+31) *V

V = 0.26 m/s

B)

To find the fraction of the energy that remains, divide the final kinetic energy by the initial kinetic energy.

Kf/Ki = 0.5( 45+31)*0.262 / ( 0.5*45*7.952 + 0.5*31*10.92)

Kf/Ki = 0.00078

C)

The conservation of momentum calculations are the same as part (a), so the center of mass of the boys is still moving to the right at 0.26 m / s after the collision.

D)

To find the angular speed of the two-boy object, RE need to find the total angular momentum of the system and the moment of inertia of the two boys after linking frith respect to the point where their hands meet). The angular momentum of a particle with respect to an axis is equal to the linear momentum of that particle times the closest distance that particle gets to that axis.

L=m*v*r

= 45*7.95*1.02/2 + 31*10.09*1.02/2 = 341.97 kg*m2/s

I = MR2 = 45*0.512 +31*0.512 = 19.76 kg*m2

now

L= I*w

w = L/ I = 341.97/19.76

w = 17.30 rad/s

E)

To find the fraction of the energy that remains, divide the final kinetic energy by the initial kinetic energy.

Kfinal(linear) + Kfinal(angular)/Kinitial = [1/2*(45+31)*0.262 + 1/2*19.76*17.302]/[1/2*45*7.952 + 1/2*31*10.092]*100

= 0.9864 *100

= 98.64 %

F)

The answers are so different because head-on collisions between similarly sized objects are grossly inefficient . if the two kids were the size and had the same velocity , conservation of momentum states that they would lose all of their kinetic energy . on the other hand , glancing blows like this one allow a lot of energy to be transfered in to rotational kinetic energy , causing much less energy to be lost on impact .


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