Question

Stranded on a frozen and frictionless lake, David, who has a mass of 55 kg, takes off his new 0.15 kg Michael Jordan tennis shoes that he got from Santa Clause for Christmas. He throws the shoes horizontally directly away from the shore with a speed of 2 m/s. If David is 5 m away from shore, how long does it take before he reaches land

Answer 1

t = 909.1 s

Step-by-step explanation:

• Assuming no other external forces acting when David threw the shoes, total momentum must be conserved.
• Before throwing the shoes, both objects were at rest, so the initial momentum was just zero.
• After throwing them, total momentum must keep being zero.
• Total final momentum is just the sum of the momentum of the shoes (going away from the shore) and David (which must be going to the shore), as we can see here:

`p_(o) =p_(f) = (m_(shoes) * v_(shoes)) +M_(Dav) * v_(D) = 0 (1)`

• From the problem, we have the following givens:
• msh = 0.15 kg
• vsh = 2.0 m/s (Assuming as positive the direction away from the shore)
• MD = 55 kg.
• So, the only unknown that remains is vD, which can be get from (1) as follows:

`v_(D) = (-(m_(shoes)*v_(shoes)))/(M_(Dav)) = (-(0.15kg*2.0m/s))/(55 kg)} = -0.0055 m/s (2)`

• assuming no friction, after throwing the shoes, due there are no external forces acting in the horizontal direction, David keeps moving at this constant speed towards the shore in a straight line.
• Applying the definition of average velocity, and knowing that he is at 5 m from land (taking as the origin the point where he was when threw the shoes, so the displacement is negative) we can get how long it takes him to reach land:

`t = (x_(f)- x_(o) )/(v_(D) ) = (-5m)/(-0.0055m/s) = 909.1 s (3)`