Question

A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.
a) What is the initial momentum of the mass?
b) What is the final momentum of the mass?
c) What was the force acting on the mass?
d) What was the impulse acting on the mass?

Answer 1

$\huge{\textbf{\underline{As per Question:-}}}$

Mass of the system = 50 kg

Let a force F acts on it for 2s until it's velocity becomes 3 m/s.

u = 0

v = 3m/s

$\huge{\textbf{\underline{Solution :-}}}$

$\large{\textbf{\underline{Momentum }}}$

Momentum is defined as the product of mass and velocity of an object.

$P_i = mu$

Since, body starts from rest so,

u = 0

$P_i = 0$

$\large{\textbf{ Initial\: momentum= 0 }}$

$P_f = mv$

$P_f = 50 \times 3$

$P_f= 150 kg m/s$

$\large{\textbf{ Final\: momentum = 150 kg m/s }}$

Since, acceleration of the system will be.

$v = u + at$

$3 = 0 + a \times 2$

$3 = 2a$

$a = \dfrac{3}{2} m/s^2$

Force acting on the mass will be :-

$F = ma$

$F = 50 \times \dfrac{3}{2}$

$F = 25 \times 3$

$F = 75N$

$\large\textbf{ Force \:acting\: on \:the \:mass\: will\: be \:75\: N }$

$\large{\textbf{\underline{Impulse:-}}}$

The force acting on an object for instant time is said to be impulse.

$I = F T$

$I = 75 \times 2$

$I = 150 Ns$

$\large\textbf{ Impulse\: acting\: on\:mass = 150 Ns }$

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the body (m) = 50 Kg.
• Time period (t) = 2 seconds.
• Velocity (v) = 3 m/s.
• Initial velocity (u) = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

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By momentum formula,

$\large{\boxed{\tt P = mv}}$

For initial momentum,

$\large{\boxed{\tt P_i = mu}}$

Substituting the values,

$\large{\tt P_i = 50 \times 0}$

As initial velocity is Zero.

$\huge{\boxed{\boxed{\tt P_i = 0 \: Kgm/s}}}$

So, the initial momentum is 0 Kgm/s.

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By momentum formula,

$\large{\boxed{\tt P = mv}}$

For Final momentum,

$\large{\boxed{\tt P_f = mv}}$

Substituting the values,

$\large{\tt P_f = 50 \times 3}$

As final velocity is 3 m/s.

$\huge{\boxed{\boxed{\tt P_f = 150 \: Kgm/s}}}$

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As we know,

The rate of change of momentum is directly proportional to The force applied.

Now,

$\large{\boxed{\tt F = \dfrac{ \Delta P}{t}}}$

Now,

$\large{\tt F = \dfrac{ P_f - P_i}{t}}$

Substituting the values,

$\large{\tt F = \dfrac{150 - 0}{2}}$

$\large{\tt F = \dfrac{150}{2}}$

$\large{\tt F = \dfrac{\cancel{150}}{\cancel{2}}}$

$\huge{\boxed{\boxed{\tt F = 75 \: N}}}$

So, the Force acting on mass is 50 N.

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As we know, From Impulse formula,

$\large{\boxed{\tt I = F \times t}}$

Substituting the values,

$\large{\tt I = 75 \times 2}$

$\huge{\boxed{\boxed{\tt I = 150 \: Kgm/s}}}$

So, the Impulse imparted is 150 Kgm/s .

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