Question

A body of mass $2\text{kg}$ has kinetic energy of $400\text{J}$. A constant force of $10\text{N}$ is applied in the opposite direction of velocity. Find the magnitude of the momentum of the body $2\text{s}$ after application of the force.

• A. $40\text{kg m/s}$
• B. $60\text{kg m/s}$
• C. $30\text{kg m/s}$
• D. $20\text{kg m/s}$

Answer 1

The correct option is D $20\text{kg m/s}$

Let the initial velocity of the object is $u_0$
So,
$K.E.=\frac{1}{2}m{u}_0^2$
$400=\frac{1}{2}\times 2\times u_0^2$
$u_0=\sqrt{400}$
$u_0=20\text{m/s}$
For finding the acceleration of body due to opposite force $F=10\text{N}$,
$-F=ma$ [Force is in opposite direction]
$-10=2\times a$
$a=-5{\text{m/s}}^2$
Applying $1^{st}$ equation of motion
$v=u_0+at$
$v=20+(-5)\times 2$
$v=10\text{m/s}$
$\therefore$The momentum $P$ is
$P=mv=10\times 2=20\text{kg m/s}$