Chapter 9.1, Problem 3PP

To determine

The resulting speed and direction of motion of the object after each impulse.

Answer to Problem 3PP

When the force applied on an object is $F=5\text{ N}$ , the resulting speed is $2.7\text{ m/s}$ and is in the same direction as the initial velocity.

When the force applied on an object is $F=-5\text{ N}$ , the resulting speed is $\text{1}\text{.28 m/s}$ and is in the same direction as the initial velocity.

Explanation of Solution

Given:

Mass of an object is $m=7.0\text{ kg}$

Velocity of an object is $v\text{ = 2}\text{.0 m/s}$

An object receives two impulses (one after the other) along the direction of its motion. The two impulses are:

Impulse 1:

Force applied on an object is $F=5\text{ N}$ or $5\text{ kg}\cdot {\text{m/s}}^2$ and the time interval is $\Delta t=1\text{ s}$ .

Impulse 2:

Force applied on an object is $F=-5\text{ N}$ or $-5\text{ kg}\cdot {\text{m/s}}^2$ and the time interval is $\Delta t=1\text{ s}$ .

Formula used:

Impulse experienced by an object must be the change in momentum of that object. Thisis given by the expression,

  $F\Delta t=p_f-p_i$$\left(1\right)$

Where,

  $F$ is force on the object,

  $\Delta t$ is the time interval

  $p_f$ is the final momentum of an object, $p_f=m{v}_f$

  $p_i$ is the initial momentum of an object, $p_i=m{v}_i$

  $m$ is mass of an object,

  $v_i$ and $v_f$ are the initial and final velocity of an object

Substituting for $p_f$ and $p_i$ in equation $\left(1\right)$ ,

  $F\Delta t=m{v}_f-m{v}_i$

  $\Rightarrow v_f=\frac{F\Delta t+m{v}_i}{m}$$\left(2\right)$

Calculation:

When the force applied on an object is $F=5\text{ N}$ or $5\text{ kg}\cdot {\text{m/s}}^2$ , resulting speed is obtained by substituting the numerical values in equation $\left(2\right)$ ,

  $v_f=\frac{\left(5\text{ kg}\cdot {\text{m/s}}^2\right)\left(1\text{ s}\right)+\left(7.0\text{ kg}\right)\left(\text{2}\text{.0 m/s}\right)}{\left(7.0\text{ kg}\right)}$

  $=\frac{5\text{ kg}\cdot {\text{m/s}}^2\text{ + 14 kg}\text{.m/s}}{7.0\text{ kg}}$

  $=\frac{19\text{ kg}\cdot {\text{m/s}}^2}{7\text{ kg}}=2.7\text{ m/s}$ , and this final velocity is in the same direction as the initial velocity.

When the force applied on an object is $F=-5\text{ N}$ or $-5\text{ kg}{\text{.m/s}}^2$ , resulting speed is obtained by substituting the numerical values in equation $\left(2\right)$ ,

  $v_f=\frac{\left(-5\text{ kg}\cdot {\text{m/s}}^2\right)\left(1\text{ s}\right)+\left(7.0\text{ kg}\right)\left(\text{2}\text{.0 m/s}\right)}{\left(7.0\text{ kg}\right)}$

  $=\frac{-5\text{ kg}\cdot {\text{m/s}}^2\text{ + 14 kg}\cdot {\text{m/s}}^2}{7.0\text{ kg}}$

  $=\frac{9\text{ kg}\cdot {\text{m/s}}^2}{7\text{ kg}}=1.28\text{ m/s}$ , and this final velocity is in the same direction as the initial velocity.

Conclusion:

When the force applied on an object is $F=5\text{ N}$ , the resulting speed is $2.7\text{ m/s}$ and is in the same direction as the initial velocity.

When the force applied on an object is $F=-5\text{ N}$ , the resulting speed is $\text{1}\text{.28 m/s}$ and is in the same direction as the initial velocity.