mass 1 kg initially at rest is given. What will be the graph of momentum Vs time for the given graph?

Options

A.
B.
C.
D.

Answer

A

Explanation

`triangle vecP=int vecF dt`
for `tle0.05sec" "F=5/0.05t`
`impliesF=100t`
for `0.05 sec le t le 0.1 sec" "F=-100t+C`
at `t=0.1 sec F=0" ""so",`
`0=-10+C" "implies C=10`
for 0 to 0.05 sec or 500 m/s
`P_2-P_1=overset(0.05)underset(0)int100t dt" "implies P_2`
`=100(t^2/2)_(0)^(0.05)`
`P_2=50 xx (1/20)^2" "impliesP_2`
`=(50)/400`
`P_2=1/8" "P_2`
`=0.125 "N.S"`
For 0.05 sec to 0.1 sec
`P_2-0.125=overset(0.1)underset(0.05)int(-100t+10)dt`
`implies P_2-0.125=-100(T^2/2)_(0.05)^(0.1)+10(t)_(0.05)^(0.1)`
`P_2=0.125=50[1/100-1/400]+10[1/10-1/20]`
`implies P_2=0.125=-50[(4-1)/400]+1/2`
`P_2-0.125=-50 xx 3/400+1/2`
`P_2-0.125=-3/8+1/2`
`P_2-0.125=(-3+4)/8`
`P_2-0.125=0.125" "implies" "P_2`
`=0.25 kg-m//s`