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 8.03 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.0 m/s along the same line. When they meet, they grab each other and hang on. (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?
I really need help ASAP
Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 8.03 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.0 m/s along the same line. When they meet, they grab each other and hang on. (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? Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 8.03 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.0 m/s along the same line. When they meet, they grab each other and hang on. (a) What is their velocity immediately thereafter? magnitude 0.675 m/s direction right (b) What fraction of their original kinetic energy mechanical energy after their collision? .0057 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.% (c) That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.07 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 0.675 m/s direction I right (d) Find their angular speed. 2.6141 X Your response differs from the correct answer by more than 10%. Double check your calculations. 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?
Explanation / Answer
a) m1*v1 - m2*v2 = (m1+m2)*V
V = ((45*8.03)-(31*10))/(45+31)
V = + 0.675 m/s
direction right
b)KEf = 0.5*(m1+m2)*v^2 = 17.31375 J
KEi = 0.5*m1*v1^2 + 0.5*m2*v2^2 = 3000.82025 J
KEi = (0.5*45*8.03*8.03)+(0.5*31*10*10)
fraction = KEf/KEi = 0.58 %
c)
m1*v1 - m2*v2 = (m1+m2)*V
V = ((45*8.03)-(31*10))/(45+31)
V = + 0.675 m/s
direction right
d) w = v/r = 0.675/(1.07/2) = 1.2618 rad/s
e)
KEi = 0.5*m1*v1^2 + 0.5*m2*v2^2 = 3000.82025 J
KEi = (0.5*45*8.03*8.03)+(0.5*31*10*10)
KEf = 0.5*(I1+I2)*w^2
KEf = 0.5*((45*.535*.535)+(31*.435*.535))*1.2618^2
KEf = 15.997 J
fraction = 0.53 %
f) in e it is rotatory motion
