SOLUTION: A 8.80 kg object with a speed of 33.0 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebounds at the same speed and angle on the other side

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Question 1057837: A 8.80 kg object with a speed of 33.0 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebounds at the same speed and angle on the other side of the normal. What is the change (in magnitude only) of the linear momentum of the object (in kilogram-meters/second)?
Answer by ikleyn(52884)   (Show Source): You can put this solution on YOUR website!
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A 8.80 kg object with a speed of 33.0 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate,
and rebounds at the same speed and angle on the other side of the normal. What is the change (in magnitude only)
of the linear momentum of the object (in kilogram-meters/second)?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The initial momentum is the vector of the magnitude 8.8*33 = 290.4 (kg*m/s) directed coherently with the vector of initial speed.

The momentum after the strike has THE SAME magnitude, but is directed perpendicularly to the vector of the initial momentum.

So, you have two momentum vectors of the known magnitude perpendicular each other.

The difference of these two vectors is the hypotenuse of a right angled triangle with the sides 290.4.

Can you calculate the magnitude of this vector?

It is  =  (kg*m/s).

Answer. (kg*m/s).


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