Finishing_class

course Phy 241

How much time do I have to complete the class? I will be taking test1 at the beginning of next week.

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Student Name:

assignment #017

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21:39:31

`q001. Note that this assignment contains 5 questions.

. A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.

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RESPONSE -->

F=(2kg)(3m/s)/.03s=200N

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21:40:34

By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N.

Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.

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RESPONSE -->

I calculated the force for the 2kg object.

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21:40:48

`q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.

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RESPONSE -->

200N

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21:41:43

Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second.

This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object.

A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s.

Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.

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RESPONSE -->

Ok so you cannot assume that the change in velocity is directly shifted to the other object.

That's not quite correct. The forces are equal and opposite. If the masses are different, the accelerations and therefore the changes in velocity of the two objects will therefore be different.

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21:42:32

`q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?

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Ignoring the friction it should be equal

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21:44:45

The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules.

The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules.

Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic) is involved.

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RESPONSE -->

ok, the equations to figure this out makes sense, but where does the energy get transfered to?

Energy can be dissipated in the form of thermal energy (heat), light, sound; energy can also be converted to chemical or even nuclear potential energy.

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21:48:33

`q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?

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I know that momentum is m`dv, but there are two masses and two `dv. So if I take the 10kg object, we have 10kg *-3m/s. -30 momentum. If we have 2kg*10 we have 20 kg*m/s. Then the total would be 10 kg*m/s.

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21:52:07

The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision.

The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second.

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

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RESPONSE -->

ok so momentum is m*v, so the total is only 50 because v is 0 for the second object because it is at rest. I am not sure why we do not have -30 as the the momentum for the first object after collsion because the change in v is negative.

Momentum is m v. If you have a lot of momentum in the positive direction, and then have a relatively small negative change in momentum, the momentum will still be positive. For example if your car hits a deer, it loses momentum but it will still be going forward.

Here you use `dv instead of just v which would be 2m/s.

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21:54:30

`q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?

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All energy must be the same if there is no friction, and momentum is equal to the Impulse, F`dt.

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21:55:00

Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum.

Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.

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RESPONSE -->

ok

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Good job. See my notes and let me know if you have questions.