Question

A force of 10N acts on a body of mass 2kg for 3s, initially at rest. Find its velocity acquired by the body and change in momentum of the body?

Answer 1

We know F=ma
a=F/m=10/2=5 m/s²
Since the body is intially at Rest,u=0
By 1st equation of motion v= u +at
Here t=3s
u=0
So,v=5×3=15 m/s

Answer 2

Answer:

A force of 10N acts on a body of mass 2kg for 3s, initially at rest.

• Its velocity acquired by the body is $15\ m/s$

• Change in the momentum of the body is $30\ kgm/s$

Step-by-step explanation:

The first equation motion is given by,

$v = u+at$                   ...(1)

Where,

u -  Initial velocity

v -  Final velocity

t  -  Time

a -  Acceleration

Acceleration is the ratio of change in velocity and time.

That is,   $a = \frac{v-u}{t}$        ...(2)

Newton's second law of motion is given by the expression,

$F = ma$

Where

F - Force

m - Mass of the body

Substitute equation(2) in (1)           

$F = m(\frac{v-u}{t})$                  ...(3)

The momentum (P) of the body is given by,

$Momentum (P) = Mass (m)\times Velocity (v)$

Or $P = mv$

Change in momentum

$\Delta P = m(\Delta v)= m(\ v -u)$       ...(4)

       

From the question,

$F = 10\ N$

$m = 2\ kg$

$u = 0\ m/s$

$t = 3\ s$

Substitute these values into equation (3)

$10 = 2(\frac{v-0}{3})$

$\frac{10}{2}= \frac{v}{3}$

$v = 15\ m/s$

Final velocity = $15\ m/s$

Substitute all these values into equation (4)

$\Delta P = 2\times (15 - 0)$

       $= 30 \ kgm/s$