A 738 N student stands in the middle of a frozen pond having a radius of 5.6 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 2.9 kg physics textbook horizontally toward the north shore at a speed of 4.6 m/s. How long does it take him to reach the south shore?
________ s
________ s
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Conservation of momentum
m1v1,1 + m2v2,1 = m1v1,2 + m2v2,2
v1,1 is the velocity of object 1 at time 1. v1,2 is the velocity of object 1 and time 2.
Since neither object is moving to start with both v1,1 and v2,1 are zero. This give
m1v1,2 + m2v2,2 = 0
m1v1,2 = - m2v2,2
If m1 is the mass of the student:
m1 = F/g = 738 N / 9.81m/s^2 = 75.23 kg
Now complete the momentum equation:
75.23 kg* v1,2 = - 2.9 kg * 4.6 m/s
v1,2 = -2.9*4.6/75.23 = -0.1773 m/s <---- students velocity in opposite direction of book
Now, to find time.
t = 5.6 m / 0.1773 m/s = 31.6 seconds
m1v1,1 + m2v2,1 = m1v1,2 + m2v2,2
v1,1 is the velocity of object 1 at time 1. v1,2 is the velocity of object 1 and time 2.
Since neither object is moving to start with both v1,1 and v2,1 are zero. This give
m1v1,2 + m2v2,2 = 0
m1v1,2 = - m2v2,2
If m1 is the mass of the student:
m1 = F/g = 738 N / 9.81m/s^2 = 75.23 kg
Now complete the momentum equation:
75.23 kg* v1,2 = - 2.9 kg * 4.6 m/s
v1,2 = -2.9*4.6/75.23 = -0.1773 m/s <---- students velocity in opposite direction of book
Now, to find time.
t = 5.6 m / 0.1773 m/s = 31.6 seconds
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um, 33.