Physics - drwls

Posted by winterWX on Monday, March 19, 2007 at 11:51pm.

A 2600 kg truck traveling north at 38 km/hr turns east and accelerates to 55 km/hr. What is the change in the truck's kinetic energy? What is the magnitude of the change in the linear momentum of the truck? What is the direction of the change in linear momentum of the truck? (give an angle, measured from east)

For Further Reading

* Physics - drwls? - drwls, Tuesday, March 20, 2007 at 2:17am

Convert the speeds to m/s and use (1/2) MV^2 for the kinetic energy. Momentum is MV, but is a vector so you must account for direction changes when calculating momentum change.

You need to learn the basic concepts and try applying them yourself. We will be happy to check your work.

-----------------------

38 km/hr = 10.6 m/s
55 km/hr = 15.3 m/s

.. as for finding the magnitude in the linear momentum of the truck.. im having trouble figuring that out.. must i use the Scalar Product? and put the north and the east directions into vectors?

such as..

a x b = ab cos(theta)

(10.6cos[90])(15.3) x (15.3cos[90])(10.6) ..?

.. the direction change in an angle would be 90 degrees right?

i feel like im overanalyzing this..

The kinetic energy changes from
(1/2)(2600)(10.6)^2 = ? Joules to
(1/2)(2600)(15.3)^2 = ? Joules
Do the numbers.
The initial momentum is 2600*10.6 = ? kg m/s North
The final momentum is 2600*15.3 = ? kg m/s East
The angle of the direction CHANGE is arctan 38/55 south of east. (Draw a right momentum triangle to convince yourself of that). The magnitude of the direction change vector is the square root of the sum of the squares of the initial and final momenta, since they are at right angles to one another.

gotcha. thanks a bunch :)

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