If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

022. `query 22

Question `gen, `prin (Optional Openstax):

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?

Your solution:

Confidence rating::

Given Solution:

The bears exert equal and opposite forces on one another, which act for equal time intervals.  If no forces act other than these contact forces, this results in equal and opposite changes in momentum.

The collision takes place on ice, so frictional forces will be small and for the short interval of the collision may be disregarded.  The total momentum of the two-bear system therefore remains constant during the collision.

The total momentum after collision is therefore equal to the total momentum before.

Before collision the second bear is stationary, so the total momentum is just the (.2 kg * .75 m/s) = .15 kg m/s momentum of the first bear.

After collision the two bears constitute a mass of .2 kg + .35 kg = .55 kg, and we do not yet know their common velocity. 

If we let u stand for their common unknown velocity, their momentum after collision is therefore

momentum after collision = .55 kg * u.

Since the momentum is the same after collision as it was before, we therefore have

.15 kg m/s = .55 kg * u

so that

u = .15 kg m/s / (.55 kg) = .7 m/s, approximately.

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Question: `qQuery gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west

Your solution:

Confidence rating::

Given Solution:

`a** We'll take East to be the positive direction.

The original magnitude and direction of the momentum of the fullback is

p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East.

The magnitude and direction of the impulse exerted on the fullback will therefore be

impulse = change in momentum or

impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s.

Impulse is negative so the direction is in the negative x direction, i.e., West.

Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is

Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N

The direction is in the negative x direction, i.e., West.

The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N.

The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0).

The impulse on the tackler is to the East. **

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