Question

A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs 720 N and the shoe weighs 4.0 N, the man moves to the left at approximately______.A) 0B) 2.1 x 102 m/s.C) 8.3 x 102 m/s.D) 15 m/s.E) 2.7 x 10-3 m/s 19.

Answer 1

The man moves to the left at approximately 0 m/s.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The man and the shoe are initially at rest on the frictionless ice, so the initial momentum is zero. When the man throws the shoe to the right, it gains momentum in that direction. According to the conservation of momentum, the man must gain an equal and opposite momentum in the left direction. Since the man weighs more than the shoe, his velocity will be significantly less than the shoe's velocity. Therefore, the man moves to the left at approximately 0 m/s.

Answer 2

A 0.083 m/s approx 0

Step-by-step explanation:

mass of the man is 720 N, the mass of the shoe is 4 N, and the man and the shoe were initially at rest. After throwing the shoe, the shoe had a velocity of 15 m/s. Using conservation of momentum:

since the man and the shoe were initially at rest, their initial momentum is zero

0 = M1V1 + M2V2 where M1 is the mass of the shoe ( 4 / 9.81), V1 is the velocity of the shoe 15m/s, M2 is mass of the man ( 720 / 9.81), V2 is the velocity of the man

MAKE V2 subject of the formula

- M1V1 = M2V2

- M1V1 / M2

substitute the values into the equation

- ((4/9.8) × 15) /( 720 / 9.81) = V2

V2 = - 0.0833 m/s approx 0