A constant force F acts for 1 sec, on a body of mass 1kg moving with perpendicular to its initial velocity, then

Question

A constant force F acts for 1 sec, on a body of mass 1kg moving with perpendicular to its initial velocity, then

A constant force F acts for 1 sec, on a body of mass 1kg moving with perpendicular to its initial velocity, then
A. distance covered by the body is `(u+(F)/(2))`
B. displacement of the body is `sqrt(u^(2)+((F)/(2))^(2))`
C. change in kinetic energy of the body is `sqrt((1)/(2)(u^(2)+F^(2)))`
D. momentum of the body is increased by F/2.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

m=1kg
`s_(y)=0+(1)/(2)at^(2)`
`s_(y)=(1)/(2)((F)/(1))(I)^(2)=(F)/(2)`
t=1sec, `s_(x)=u`
`s=sqrt((s_(x))^(2)+(s_(y))^(2))=sqrt(u^(2)+((F)/(2))^(2))`
`W_(F)=DeltaK`
`Fxx(F)/(2)=(1)/(2)mV^(2)-(1)/(2)mu^(2)`
`(F^(2))/(2)+(1)/(2)u^(2)=(V^(2))/(2)`
`V=sqrt(F^(2)+u^(2))`
J=`DeltaP`
`FxxI=DeltaP=F`
As constant force acts on the body and angle between u `&` F is neither `0^(@)` nor `180^(@)`, path is parabolic.