Chapter 9, Problem 93A

a.

To determine

Momentum of dancer.

Answer to Problem 93A

Momentum of dancer, $P=150.26\text{ kg-m/s}$

Explanation of Solution

Given:

Mass of dancer, $m=60\text{ kg}$

Height, $h=0.32\text{ m}$

Formula used:

According to third equation of motion.

  $v^2=u^2+2as$

Here, v is final velocity, u is initial velocity, a is acceleration and s is distance.

  $P=mv$

Here, P is momentum, m is mass and v is velocity.

Calculation:

Initial velocity of dancer, $u=0\text{ m/s}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Now, substitute $a=g$ and $s=h$ , u and solve for v.

  $\begin{array}{l}{v}^2=0^2+2\times 9.8\times 0.32\\ v^2=6.272\\ v=2.504\text{ m/s}\end{array}$

Now, substitute the value of m and v and solve.

  $\begin{array}{c}P=60\times 2.504\\ =150.26\text{ kg-m/s}\end{array}$

Conclusion:

Therefore, momentum of dancer is $150.26\text{ kg-m/s}$ .

b.

To determine

Impulse needed to stop the dancer.

Answer to Problem 93A

Impulse, $J=150.26\text{ N-s}$

Explanation of Solution

Given:

Mass of dancer, $m=60\text{ kg}$

Height, $h=0.32\text{ m}$

Formula used:

Impulse of a body is equal to change in momentum.

  $J=m\left(v-u\right)$

Here, J is impulse, v is final velocity and u is initial velocity.

Calculation:

From part (a) $v=2.504\text{ m/s}$

Substitute the value of m, v and u and solve.

  $\begin{array}{l}J=60\left(2.504-0\right)\\ =150.26\text{ N-s}\end{array}$

Conclusion:

Therefore, impulse need to stop the dancer is $150.26\text{ N-s}$ .

c.

To determine

Average force exerted on body of dancer.

Answer to Problem 93A

Average force, $F=3005.2\text{ N}$

Explanation of Solution

Given:

Mass of dancer, $m=60\text{ kg}$

Height, $h=0.32\text{ m}$

Time interval, $\Delta t=0.050\text{ s}$

Formula used:

Impulse of a body is equal to change in momentum.

  $F\Delta t=m\left(v-u\right)$

Here, F is force, $\Delta t$ is time interval, v is final velocity and u is initial velocity.

Calculation:

From part (a) $v=2.504\text{ m/s}$

Substitute the value of m, v, u and $\Delta t$ and solve.

  $\begin{array}{l}F\times 0.050=60\left(2.504-0\right)\\ F\times 0.050=150.26\\ F=\frac{150.26}{0.050}\\ =3005.2\text{ N}\end{array}$

Conclusion:

Therefore, average force on body of dancer is 3005.2 N.

d.

To determine

Comparison of stopping force and weight of dancer.

Answer to Problem 93A

Stopping force is greater than weight.

Explanation of Solution

Given :

Mass of dancer, $m=60\text{ kg}$

Height, $h=0.32\text{ m}$

Time interval, $\Delta t=0.050\text{ s}$

Formula used :

  $W=mg$

Here, W is weight, m is mass and g is acceleration due to gravity.

Calculation :

From part (c) $F=3005.2\text{ N}$

Substitute the value of m, and g and solve.

  $\begin{array}{l}W=mg\\ =60\times 9.8\\ =588\text{ N}\end{array}$

Conclusion :

Therefore, stopping force is greater than weight of dancer.