Question

A force F acts for 1 sec on a body of mass 1kg moving with initial velocity u in a direction perpendicular to its initial velocity

• A. distance covered by the body is $(u+\frac{F}{2})$
• B. displacement of the body is $\sqrt{u^2+(F/2{)}^2}$
• C. change in kinetic energy of the body is $\sqrt{\left(1/2\right)(u^2+F^2})$
• D. momentum of the body is increased by F/2.

Answer 1

Answer: (B)

displacement of the body is $\sqrt{u^2+(F/2{)}^2}$

m = 1 kg
$s_y=0+\frac{1}{2}a{t}^2$
$s_y=\frac{1}{2}\left(\frac{F}{1}\right)(1{)}^2=\frac{F}{2}$
t = 1 sec, $s_x=u$
$s=\sqrt{(s_x{)}^2+(s_y{)}^2}=\sqrt{u^2+(F/2{)}^2}$
$W_F=\Delta K$
$F\times F/2=1/2m{v}^2-1/2m{u}^2$
$F^2/2+1/2u^2=V^2/2$
$V=\sqrt{F^2+u^2}$
$J=\Delta P$
$F\times 1=\Delta P=F$
As constant force acts on the body and angle between u & F is niether $o^0$ nor ${180}^0$, path is parabolic.