Asked by JasonJones 11 months ago

A 5 kg steel ball strikes a wall with a speed of 15 m/s at an angle of 30o with the normal to the wall. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.2 s, what is the magnitude of the average force exerted on the ball by the wall?

Answered by lancer 11 months ago

$ I=\int\limits_{t_1}^{t_2}dt $

also, when the force and mass are constant then:

$ I=F \Delta t=m \Delta v = \Delta p $

where

$ F $ = constant total net force applied

$ \Delta t $ = time interval over which the force is applied

$ m $ = mass of the object

$ \Delta v $ = change in velocity produced by the force

given

$ M $ = 3 kg

$ v $ = 15 m/s

$ \theta $ = 30o

The y component of the momentum is not changed in this case. The x component is changed by an amount:

$ \Delta P_x =-2Mv cos(\theta) $

and

$ F \Delta t = \Delta p \,\,\to\,\, F=\frac{\Delta p}{\Delta t} = \frac{-2Mv cos(\theta)}{\Delta t} = \frac{-2(5\,kg)(15\,m/s)cos(30^0)}{0.2\,s}=-649.5\,N $

$ ||F||=649.5 N $

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Asked: 11 months ago