Question

A force of 10 N acts on a body of mass 2 kg for 3s, initially at rest. Calculate:
(i) The velocity acquired by the body
(ii) Change in momentum of the body

Answer 1

Calculate the acceleration produced in the body using Newton’s second law. Use the first kinematic equation to determine the final velocity of the body after 3 s. The momentum of the body is the product of its mass and its velocity.

Formula used:
Newton’s second law, $F = ma$,
where, m is the mass and a is the acceleration.
First kinematic equation,$v = u + at$,
where, u is the initial velocity, v is the final velocity and t is the time.
Momentum, $p = mv$,
where v is the velocity.

Complete step by step answer:
We have given the force acting on the body $F = 10\,{\text{N}}$ and the mass of the body is $m = 2\,{\text{kg}}$.
(i) We can determine the acceleration of the body using Newton’s second law of motion as follows, $a = \dfrac{F}{m}$
Here, F is the force and m is the mass of the body.
Substituting $F = 10\,{\text{N}}$ and $m = 2\,{\text{kg}}$ in the above equation, we get,
$a = \dfrac{{10}}{2}$
$\Rightarrow a = 5\,{\text{m/}}{{\text{s}}^2}$
Let’s determine the final velocity of the body using the first kinematic equation as follows,
$v = u + at$
Here, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time.
Substituting $u = 0$, $a = 5\,{\text{m/}}{{\text{s}}^2}$ and $t = 3\,{\text{s}}$ in the above equation, we get,
$v = 0 + \left( 5 \right)\left( 3 \right)$
$\therefore v = 15\,{\text{m/s}}$

Therefore, the final velocity of the body is 15 m/s.

(ii) We can express the change in the momentum of the body as,
$\Delta p = mv - mu$
$\Rightarrow \Delta p = m\left( {v - u} \right)$
Substituting $m = 2\,{\text{kg}}$, $v = 15\,{\text{m/s}}$ and $u = 0$ in the above equation, we get,
$\Delta p = \left( 2 \right)\left( {15 - 0} \right)$
$\therefore \Delta p = 30\,{\text{kg}}\,{\text{m/s}}$

Therefore, the change in the momentum of the body is 30 kg m/s.

Note: Students can determine the change in momentum of body using Newton’s second law of motion, $F = \dfrac{{dp}}{{dt}}$. Substituting 10 N for force and 3 s for time, we can get the change in momentum 30 kg m/s. We have taken the initial velocity as zero since the body was initially at rest.