Question

The magnitude of the net force exerted in the x direction on a 3.90-kg particle varies in time as shown in the figure below.

(a) Find the impulse of the force over the 5.00-s time interval. I = ? (N · s)

(b) Find the final velocity the particle attains if it is originally at rest. Vf = ? (m/s)

(c) Find its final velocity if its original velocity is -3.10 i m/s. Vf = ? (m/s)

(d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s. Favg = ? (N)

Answer 1

Concepts and reason

The concepts used to solve this question is the impulse

Initially, draw the graph and then the impulse is determined by calculating the area under the graph. Later use the impulse to find the initial and final velocities of the particle. At last use the relation between impulse, force and time to determine average force acting on the particle.

Fundamentals

Impulse:

Impulse is a vector quantity and has both magnitudes as well as direction. It is defined as the product of the force acting on the particle and time interval at which force is acting on the particle.

The magnitude of the impulse is,

$I = F\Delta t$

Here, $F$ is the force acting on the particle and $\Delta t$ is the time interval.

The impulse is also defined as the change in momentum of the particle.

$I = m\Delta v$

Here, $m$ is the mass of the particle and $\Delta v$ is the change in velocity.

(a)

The graph between the force and the time is drawn below.

The area of the graph gives the impulse.

Therefore,

$\begin{array}{c}\\I = Area\left( {{\rm{ABF}}} \right) + Area\left( {{\rm{BCFG}}} \right) + Area\left( {{\rm{CEG}}} \right)\\\\ = \left( {\frac{1}{2}} \right)\left( {{\rm{4 N}}} \right)\left( {2{\rm{ s}}} \right) + \left( {{\rm{4 N}}} \right)\left( {1{\rm{ s}}} \right) + \left( {\frac{1}{2}} \right)\left( {{\rm{4 N}}} \right)\left( {2{\rm{ s}}} \right)\\\\ = 2\left( {{\rm{4 N}} \cdot {\rm{s}}} \right) + 4{\rm{ N}} \cdot {\rm{s}}\\\\ = 12{\rm{ N}} \cdot {\rm{s}}\\\end{array}$

(b)

The final velocity of the particle is,

$I = m\left( {{v_{\rm{f}}} - {v_{\rm{i}}}} \right)$

The above equation is modified as,

${v_{\rm{f}}} = \frac{I}{m} + {v_{\rm{i}}}$ …… (1)

Here, ${v_{\rm{i}}}$ is the initial velocity of the particle.

Substitute $12{\rm{ N}} \cdot {\rm{s}}$ for $I$ , $3.90{\rm{ kg}}$ for $m$ and $0{\rm{ m/s}}$ for ${v_{\rm{i}}}$ in the equation (1).

$\begin{array}{c}\\{v_{\rm{f}}} = \frac{I}{m} + {v_{\rm{i}}}\\\\ = \frac{{12{\rm{ N}} \cdot {\rm{s}}}}{{3.90{\rm{ kg}}}} + 0{\rm{ m/s}}\\\\ = 3.076{\rm{ m/s}}\\\end{array}$

(c)

The final velocity of the particle is,

${v_{\rm{f}}} = \frac{I}{m} + {v_{\rm{i}}}$ …… (2)

Here, ${v_{\rm{i}}}$ is the initial velocity of the particle.

Substitute $12{\rm{ N}} \cdot {\rm{s}}$ for $I$ , $3.90{\rm{ kg}}$ for $m$ and $- {\rm{3}}{\rm{.10 m/s}}$ for ${v_{\rm{i}}}$ in the equation (2).

$\begin{array}{c}\\{v_{\rm{f}}} = \frac{I}{m} + {v_{\rm{i}}}\\\\ = \frac{{12{\rm{ N}} \cdot {\rm{s}}}}{{3.90{\rm{ kg}}}} - 3.10{\rm{ m/s}}\\\\ = - 0.024{\rm{ m/s}}\\\end{array}$

(d)

The average force exerted on the particle is,

$F = \frac{I}{{\Delta t}}$ …… (3)

Substitute $12{\rm{ N}} \cdot {\rm{s}}$ for $I$ and ${\rm{5 s}}$ for $\Delta t$ in the equation (3).

$\begin{array}{c}\\F = \frac{I}{{\Delta t}}\\\\ = \frac{{12{\rm{ N}} \cdot {\rm{s}}}}{{5{\rm{ s}}}}\\\\ = 2.4{\rm{ N}}\\\end{array}$

Ans: Part a

The impulse of the force over the time interval ${\bf{5 s}}$ is $12{\rm{ N}} \cdot {\rm{s}}$

Answer 2

Answer #1 is correct but the thing that is missing from every answer is the i hat!!!! Dont forget to put the i hat at the end of your answer!!!!