You are on a toboggan moving down a hill at 15 m/s. The toboggan has a mass of 5

Question

You are on a toboggan moving down a hill at 15 m/s. The toboggan has a mass of 5.5 kg and ss is 55 kg. If you continue on the straight path you are following, you will crash into the picture window of the chalet in front. However, 2.5 m to the left of the window is an open door. By jumping off the toboggan at right angles to its motion with a speed of 1.7 m/s, you will alter its, otherwise straight, trajectory. Calculate how far in front of the window you should jump off the toboggan and to which side you should exit.

Explanation / Answer

Tools for this problem are conservation of linear momentum, Newton's third law of motion, and similar triangles.

When I jump off the sled, I exert a force F whose reaction, negative F, sends the toboggan to the left. I need to jump off to the right.

The sled and I, combined mass of M and m, travel at a speed v and momentum (M + m)v. At separation, I have a new velocity v2f and my share of the momentum is Mv2f. The remainder is the sled's momentum, mv1f. Ignoring losses due to friction, initial momentum and final momentum are the same.

The vector sum of the sled's initial and final velocities form the legs of a right triangle whose hypotenuse makes an angle with the original path. That triangle is similar to one whose base is the distance between window and open door, and whose altitude is the distance in front of the window at which I need to jump off.

Initial momentum = (55 + 5.5)15 = 977.5 kg m/s

Final momentum = 55(1.7) + 5.5v2f = 977.5

93.5 + 5.5 v2f = 977.5

5.5v2f = 814

v2f = 148 m/s

Tan = 148/15

= arctan(148/15) = 84.2 degrees

Tan = 2.5 m/jumping off distance

jumping off distance = 2.5/tan

jumping off distance = .25 m in front of window - cutting it close.

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