What is its final velocity if its original velocity is -2.9 m/s ? What is the average force exerted on the particle for the time interval between 0 and 5.00 s.

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Answer 1 · 2018-04-13T11:55:46

(c) Impulse #vecJ=mvecv_2-mvecv_1#
Assuming initial velocity #=-2.9hati\ ms^-1#
In part (b) for initial velocity #=0# we got

#vecv_2=4.36hati\ ms^-1#

#:.#for initial velocity #=-2.9hati\ ms^-1#

Final velocity #=(4.36-2.9)#
#=1.46hati\ ms^-1#

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
Proof
Inserting various values in the formula for impulse, and using the magnitudes we get

#12=2.75v_2-2.75(-2.9)#
#=>2.75v_2=12+2.75(-2.9)#
#=>v_2=12/2.75+(-2.9)#
#=>vecv_2=1.46hati\ ms^-1#

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

(d) Impulse #J=F_"average"xx(t_2-t_1)#
Inserting given values we get

#12=F_"average"xx(5.00-0)#
#=>F_"average"=12/5=2.4\ N#