0.18 Phy1160: force and motion -- momentum, impulse, and conservation  (Page 13/14)

F = m*a, or

a = F/m = 2N/10kg = 0.2 m/s^2

A more interesting solution comes from the fact that since

impulse = F*t, and

F = m*a, then

impulse = m*a*t, or

a = impulse/(m*t) = 6*N*s/(10*kg*3*s) = 0.2 m/s^2

Pushing a wagon part 3

3. What is the velocity at the end of the 3-second interval in question 1 above.

Answer:

The impulse is equal to the change in momentum, and the initial velocity is 0.

impulse = m*(v2 - v1) = m*v2, or

v2 = impulse/m = 6*N*s/(10*kg) = 0.6 m/s

We can check that answer by knowing that the acceleration is uniform at 0.3 m/s^2 for 3 s = 0.6 m/s.

Action and reaction example

A dip in the pool

You have a body mass of 70 kg. You are on your knees on an inflatable raft in a swimming pool. The raft has a mass of 1kg. Your outstretched hands are about two meters from a safety rope that is strung across the pool.

You decide to launch yourself from the raft to catch the rope, exerting a force with a horizontal component of5 newtons. (You assume that the vertical component of your launching force will take care of the downward pull of gravity, allowing you to fly in a parabolicarc to the rope.) Assuming uniform acceleration (which is unrealistic but we will assume that anyway), how long will it take you to flythrough the air to reach the rope?

Answer:

The force that you exert on the raft will be equal and opposite to the force that the raft exerts on you. Therefore,

F = my*ay = -mr*ar

where

• my and mr are the mass of you and the raft respectively
• ay and ar are the accelerations experienced by you and the raft respectively

Therefore

ay = F/my = 5N/70kg = 0.07143 m/s^2 toward the rope

ar = -F/mr = -5N/1kg = -5.00000 m/s^2 away from the rope

We learned in an earlier module that given a constant acceleration, the distance traveled versus time is:

d = v0t + 0.5*a*t^2

In this case, v0 is zero, so

d = 0.5*ay*t^2, or

t = sqrt(d/(0.5*ay)), or

t = sqrt(2m/(0.5*0.07143m/s^2)) = 7.48324 seconds

I doubt that you will stay in the air long enough to reach the rope.

During that time period, the raft will travel the following distance in the opposite direction (assuming no resistance from the water).

d = 0.5*ar*t^2, or

d = 0.5*(-5m/s^2)*(7.48324s)^2 = -140 meters

Conservation of momentum example

Railroad cars

Getting back to my example of coupling railroad cars, when the collision has been completed, the two masses have effectively been joined into a single massand they are moving at the same velocity.

In that case, we can write the above equation as

ma*(v2 - va1) = -mb*(v2 - vb1)

ma*v2 +mb*v2 = ma*va1 +mb*vb1

v2 = (ma*va1 +mb*vb1)/(ma + mb)

where

• ma an mb represent the masses for Car-A and Car-B respectively
• v2 is the velocity of the coupled cars after the collision
• va1 and vb1 represent the initial velocities for Car-A and Car-B respectively

Then for any set of assumed mass values for the railroad cars and assumed values for the initial velocities, we can calculate the final velocity of thecoupled pair of railroad cars.

Scenario #1: Assume that the two railroad cars are just alike and empty giving them the same mass. Also assume that the initial velocity for Car-A is 10m/s and the initial velocity for Car-b is 0.