Question

A force of 12N acts on mass of 1.5 kg at rest for 3 sec .find the final velocity and the momentum of the body after 3 sec

Answer 1

F=12N m=1.5 kg t=3sec u=0(body at rest)  F=ma a=F/m a= 12/1.5 a= 8 m/s^2   According to 1st equation of motion v= u+at v= 0+(8×3) v= 24 m/s  Average velocity = u+v/2    =0+24/2    = 12 m/s   p= mv p= 1.5 × 12 p=18 kgm/s

Answer 2

Answer:

The final velocity and momentum of the body after 3sec will be equal to  24ms⁻² and 36kgms⁻¹ respectively.

Explanation:

Given: The initial velocity of the body, $u=0$

The force acting on the body, $F=12N$

Mass of the body, $m=1.5Kg$

Acceleration will be, $a=\frac{Force}{mass}$

$a=\frac{12N}{1.5Kg} =\frac{12kg\: ms^{-2}}{1.5kg}=8ms^{-2}$

From the first equation of motion:

$v=u+at$                                                     ....................(1)

Put the value of u, a and t in eq.(1):

$v=(0)+(8)(3)$

$v=24ms^{-1}$

The final velocity of the body after 3 sec is 24ms⁻².

The momentum of the body after 3sec, $p=mv$

$p=(1.5kg)(24ms^{-1})$

$p=36kgms^{-1}$

Therefore, the final momentum of the body will be 36kgms⁻¹.