Chapter 8, Problem 20A

The balls have different masses and speeds.

Rank the following from greatest to least.

a. momentum

b. the impulse needed to stop them

To determine

To rank: The momentum of the balls from greatest to least.

Answer to Problem 20A

B, D, C, A.

Explanation of Solution

Given:

The mass and velocity of four different balls are as given in the figure 1.

Mass of the ball A is $m_A=1.0\text{ kg}$

Velocity of the ball A is $v_A=9.0\text{ m/s}$

Mass of the ball B is $m_B=1.2\text{ kg}$

Velocity of the ball B is $v_B=8.5\text{ m/s}$

Mass of the ball C is $m_C=0.8\text{ kg}$

Velocity of the ball C is $v_C=12\text{ m/s}$

Mass of the ball D is $m_D=5.0\text{ kg}$

Velocity of the ball D is $v_D=2.0\text{ m/s}$

  

Formula used:

The momentum $\left(p\right)$ of an object of mass $\left(m\right)$ moving with a velocity $\left(v\right)$ is given by

  $p=mv$ $\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

The momentum of the ball A is,

  $p_A=\left(1.0\text{ kg}\right)\left(9.0\text{ m/s}\right)=9.0\text{ kg}\text{.m/s}$

The momentum of the ball B is,

  $p_B=\left(1.2\text{ kg}\right)\left(\text{8}\text{.5 m/s}\right)=10.2\text{ kg}\text{.m/s}$

The momentum of the ball C is,

  $p_C=\left(\text{0}\text{.8 kg}\right)\left(\text{12}\text{.0 m/s}\right)=9.6\text{ kg}\text{.m/s}$

The momentum of the ball D is,

  $p_D=\left(\text{5}\text{.0 kg}\right)\left(\text{2}\text{.0 m/s}\right)=10.0\text{ kg}\text{.m/s}$

Conclusion:

The rank of the balls having the momentum greatest to least is B, D, C, A.

To determine

To rank: The impulse needed to stop the balls from greatest to least.

Answer to Problem 20A

B, D, C, A.

Explanation of Solution

Given:

Mass of the ball A is $m_A=1.0\text{ kg}$

Initial velocity of the ball A is $v_{Ai}=9.0\text{ m/s}$

Final velocity of the ball A is $v_{Af}=0.0\text{ m/s}$

Mass of the ball B is $m_B=1.2\text{ kg}$

Initial velocity of the ball B is $v_{Bi}=8.5\text{ m/s}$

Final velocity of the ball B is $v_{Bf}=0.0\text{ m/s}$

Mass of the ball C is $m_C=0.8\text{ kg}$

Initial velocity of the ball C is $v_{Ci}=12\text{ m/s}$

Final velocity of the ball C is $v_{Cf}=0.0\text{ m/s}$

Mass of the ball D is $m_D=5.0\text{ kg}$

Initial velocity of the ball D is $v_{Di}=2.0\text{ m/s}$

Final velocity of the ball D is $v_{Df}=0.0\text{ m/s}$

Formula used:

Initial momentum of the ball is

  $p_i=m{v}_i$

Final momentum of the ball is

  $p_f=m{v}_f$

From the impulse - momentum theorem, the impulse needed to stop the ball can be written as

  $F\Delta t=p_f-p_i$

Substituting for $p_f$ and $p_i$ ,

  $F\Delta t=m{v}_f-m{v}_i$

  $\Rightarrow F\Delta t=m\left(v_f-v_i\right)$ $\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

The impulse needed to stop the ball A is,

  $F\Delta t=\left(1.0\text{ kg}\right)\left(\text{ 0}.0\text{ m/s}-9.0\text{ m/s}\right)=-9.0\text{ kg}\text{.m/s}$

The impulse needed to stop the ball B is,

  $F\Delta t=\left(1.2\text{ kg}\right)\left(\text{ 0}.0\text{ m/s}-8.5\text{ m/s}\right)=-\text{ 10}\text{.2 kg}\text{.m/s}$

The impulse needed to stop the ball C is,

  $F\Delta t=\left(\text{0}\text{.8 kg}\right)\left(\text{ 0}.0\text{ m/s}-12.0\text{ m/s}\right)=-\text{ 9}\text{.6 kg}\text{.m/s}$

The impulse needed to stop the ball D is,

  $F\Delta t=\left(\text{5}\text{.0 kg}\right)\left(\text{ 0}.0\text{ m/s}-2.0\text{ m/s}\right)=-\text{ 10}\text{.0 kg}\text{.m/s}$

Conclusion:

The rank of the impulse needed to stop the balls from greatest to least is B, D, C, A.