Find its vertical displacement during the time interval in which it travels 9.80

Question

Find its vertical displacement during the time interval in which it travels 9.80 cm horizontally.

Find the vertical component of its velocity after it has traveled 9.80 cm horizontally.

Find the horizontal component of its velocity after it has traveled 9.80 cm horizontally.

A proton moves at 4.50 X 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.30 X 10^3 N/C. (Ignore any gravitational effects in your calculations.) Find its vertical displacement during the time interval in which it travels 9.80 cm horizontally. Find the vertical component of its velocity after it has traveled 9.80 cm horizontally. Find the horizontal component of its velocity after it has traveled 9.80 cm horizontally.

Explanation / Answer

horizontal velocity of the proton is vh=4.5*10^5 m/s electric feild E =7.30*10^3 N/C _____________________________________________________________________________ horizontal displacement s=9.80 cm =0.098 m when the proton horizontally travelled in vertical electic feild, there is no effect in motion of proton the horizontal velocity of the proton constant through out the motion therefore velocity v =displacement / time time t =displacement /velocity           =0.098 /4.5*10^5           =2.177*10^-7 s ______________________________________________________________________________ initial vertical velocity of the proton is u =0 from the equation of motion v =u+at acceleration of the proton is a =e*E/m m is mass of the proton =1.6726*10^-27 kg e is the charge of the proton =+1.6*10^-19 C acceleration of the proton in vertical motion is   a =(1.6*10^-19)(7.3*10^3) /(1.6726*10^-27)                                                                           =6.983*10^11 m/s^2 therefore the vertical copmonent of velocity of proton after it has traveled 9.80 cm =0.098m horizontally is vv=at                          =(6.983*10^11)(2.177*10^-7)                         =15.202*10^4 m/s the horizontal component of the velocity of the proton  after it has traveled 9.80 cm =0.098m horizontally is not changed from initial value threfore horizontal component velocity of the proton is vh=4.5*10^5 m/s horizontal component velocity of the proton is vh=4.5*10^5 m/s

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