Chapter 9, Problem 45A

a.

To determine

The impulse on the puck.

Answer to Problem 45A

  $-7.13\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the hockey puck, $m=0.115\text{kg}$

Initial velocity of the puck with which it strikes the net, $v_i=37\text{}\text{m/s}$

Final velocity of the puck after it bounces off, $v_f=25\text{}\text{m/s}$ in the opposite direction

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Assume the initial direction of motion of the puck be positive, then the opposite direction will be negative. Then, the velocities will be,

  $\begin{array}{c}{v}_i=37\text{m/s}\\ v_f=-25\text{m/s}\end{array}$

Using the given values, the impulse on the puck will be,

  $\begin{array}{c}\text{Impulse}=p_f-p_i\\ =m\left(v_f-v_i\right)\\ =0.115\left(-25-37\right)\\ =-7.13\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse on the puck is $-7.13\text{kg}\cdot \text{m/s}$ .

b.

To determine

The average force on the puck if the collision takes $5.0\times {10}^{-4}\text{s}$ .

Answer to Problem 45A

  $-1.426\times {10}^4\text{N}$

Explanation of Solution

Given:

The mass of the hockey puck, $m=0.115\text{kg}$

Initial velocity of the puck with which it strikes the net, $v_i=37\text{}\text{m/s}$

Final velocity of the puck after it bounces off, $v_f=25\text{}\text{m/s}$ in the opposite direction

From part (a), the impulse on the puck is $-7.13\text{kg}\cdot \text{m/s}$ .

Time taken for the collision, $\Delta t=5.0\times {10}^{-4}\text{s}$

Formula used:

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Assume the initial direction of motion of the puck be positive, then the opposite direction will be negative.

Then the average force on the puck will be,

  $\begin{array}{c}{F}_{avg}\Delta t=\text{Impulse}\\ F_{avg}=\frac{\text{Impulse}}{\Delta t}\end{array}$

Using all the given values in above,

  $\begin{array}{c}{F}_{avg}\Delta t=\text{Impulse}\\ F_{avg}=\frac{\left(-7.13\right)}{5\times {10}^{-4}}\\ F_{avg}=-1.426\times {10}^4\text{N}\end{array}$

Conclusion:

Thus, the average force on the puck is $-1.426\times {10}^4\text{N}$

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