Resolve the Following Multiple Momentum Vectors:

1. Solution
   [latex]\vec{p} = 10 \text{ Ns West } + 10 \text{ Ns East or } 0 Ns[/latex]
2. Solution
   [latex]\vec{p} = 10 \text{ Ns East } + 10 \text{ Ns East or } 20 \text{ Ns East}[/latex]
3. Solution
   [latex]\vec{p} = 10 \text{ Ns East } + 10 \text{ Ns East } + 10 \text{ Ns East or } 30 \text{ Ns East}[/latex]
4. Solution
   [latex]\vec{p} = 10 \text{ Ns East } + 10 \text{ Ns West } + 10 \text{ Ns West or } 10 \text{ Ns}[/latex]
5. 
   Solution
   [latex]\vec{p} = 10 \text{ Ns East } + 6.0 \text{ Ns West } + 6.0 \text{ Ns West or } 2.0 \text{ Ns West}[/latex]
6. 
   Solution
   [latex]\vec{p} = 10 \text{ Ns East } + 5.0 \text{ Ns South (left like this)}[/latex]
7. 
   Solution
   Breaking the 15 Ns up, we get:
   • [latex]\vec{p_{\text{North}}} = 15 \text{ Ns } \sin 40^{\circ}[/latex] or 9.6 Ns North
   • [latex]\vec{p_{\text{East}}} = 15 \text{ Ns } \cos 40^{\circ}[/latex] or 11.5 Ns East

   [latex]\vec{p} = 11.5 \text{ Ns East } + 9.6 \text{ Ns North } + 5.0 \text{ Ns South...}[/latex]
   Resultant is [latex]11.5 \text{ Ns East } + 4.6 \text{ Ns North}[/latex]
   [latex](\approx 12 \text{ Ns East } + 4.6 \text{ Ns North}[/latex])

8.  
   Solution
   Breaking the 15 Ns up, we get:
   [latex]\begin{array} \vec{p}_{north} = 15 Ns \sin{40}^{\circ} \text{ or } 9.6 Ns \text{ North} \\ \vec{p}_{north} = 15 Ns \cos{40}^{\circ} \text{ or } 11.5 Ns \text{ East} \end{array}[/latex]Breaking the 12 Ns up, we get:
   [latex]\vec{p}_{north} = 12 Ns \sin{50}^{\circ} \text{ or } 9.2 Ns \text{ North}[/latex]
   [latex]\vec{p}_{north} = 12 Ns \cos{50}^{\circ} \text{ or } 17.7 Ns \text{ East}[/latex][latex]\vec{p}_{north} = 9.6 Ns \text{ North } + 9.2 Ns \text{ North } + 5 Ns \text{ South or } 13.8 Ns \text{ North}[/latex]
   [latex]\vec{p}_{north} = 11.5 Ns \text{ East} + 7.7 Ns \text{ West or } 4.8 Ns \text{ East}[/latex]Resultant is [latex]\vec{p} = 13.8 Ns \text{ North} 3.8 Ns \text{ East}[/latex], (≈ [latex]14 Ns \text{ North } + 3.8 Ns { East}[/latex])

A circus cannon on wheels (500 kg) launches a clown of mass 65 kg with a horizontal velocity of 4.2 m/s, calculate:

1.  Momentum of the clown (before and after)
2. The combined momentum of the clown and circus cannon before and after firing
3. The speed that this circus cannon recoils at if originally at rest

Solution

Data:

• [latex]m_{cannon} = 500 \text{ kg}[/latex]
• [latex]\vec{v}_{i\ cannon} = 0 \text{ m/s}[/latex]
• [latex]\vec{f}_{f\ cannon} = \text{ Unknown}[/latex]
• [latex]m_{clown} = 65 \text{ kg}[/latex]
• [latex]\vec{v}_{i\ clown} = 0 \text{ m/s}[/latex]
• [latex]\vec{v}_{f\ clown} = 4.2 \text{ m/s}[/latex]

Question (i):

• [latex]\vec{p}_{\text{i clown}} = (65 \text{ kg})(0 \text{ m/s}) \text{ or } 0 \text{ Ns}[/latex]
• [latex]\vec{p}_{\text{f clown}} = (65 \text{ kg})(4.2 \text{ m/s}) \text{ or } 273 \text{ Ns }(\approx 270 \text{ Ns}[/latex])

Question (ii):

• [latex]\vec{p}_{\text{i clown}} + \vec{p}_{\text{i cannon}} = 0 \text{ Ns}[/latex]
• Therefore, [latex]\vec{p}_{\text{f clown}} + \vec{p}_{\text{f cannon}} = 0 \text{ Ns}[/latex]

Question (iii):

• [latex]\vec{p}_{\text{i clown}} + \vec{p}_{\text{i cannon}} = \vec{p}_{\text{f clown}} + \vec{p}_{\text{f cannon}}[/latex]
• [latex](m_{\text{clown}})(\vec{v}_{\text{i clown}}) + (m_{\text{cannon}})(\vec{v}_{\text{i cannon}}) = (m_{\text{clown}})(\vec{v}_{\text{f clown}}) + (m_{\text{cannon}})(\vec{v}_{\text{f cannon}})[/latex]
• [latex](65 \text{ kg})(0 \text{ m/s}) + (500 \text{ kg})(0 \text{ m/s}) = (65 \text{ kg})(4.2 \text{ m/s}) + (500 \text{ kg})(\vec{v}_{\text{f cannon}})[/latex]
• [latex]0 \text{ Ns } + 0 \text{ Ns } = 273 \text{ Ns } + (500 \text{ kg})(\vec{v}_{\text{f cannon}})[/latex]
• [latex]\vec{v}_{\text{f cannon}} = -273 \text{ Ns } \div 500 \text{ kg or } -0.55 \text{m/s (cannon recoils backwards)}[/latex]

A 4.0 kg curling rock moving 0.5 m/s West collides with a stationary 4.0 kg curling rock at rest. If the original curling rock comes to a full stop upon collision what velocity should the second curling rock be moving at?

Solution

Data:

• [latex]m_{\text{rock 1}} = 4.0 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i rock 1}} = 0.5 \text{ m/s West}[/latex]
• [latex]\vec{v}_{\text{f rock 1}} = 0 \text{ m/s}[/latex]

• [latex]m_{\text{rock 2}} = 4.0 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i rock 2}} = 0.5 \text{ m/s West}[/latex]
• [latex]\vec{v}_{\text{f rock 2}} = 0 \text{ m/s}[/latex]

Solution:

• [latex]\vec{p}_{\text{i rock 1}} + \vec{p}_{\text{i rock 2}} = \vec{p}_{\text{f rock 1}} + \vec{p}_{\text{f rock 1}}[/latex]
• [latex](m_{\text{rock 1}})(\vec{v}_{\text{i rock 1}}) + (m_{\text{rock 2}})(\vec{v}_{\text{i rock 2}}) = (m_{\text{rock 1}})(\vec{v}_{\text{f rock 1}}) + (m_{\text{rock 2}})(\vec{v}_{\text{f rock 2}})[/latex]
• [latex](4.0 \text{ kg})(0.5 \text{ m/s West}) + (4.0 \text{ kg})(0 \text{ m/s}) = (4.0 \text{ kg})(0 \text{ m/s}) + (4.0 \text{ kg})(\vec{v}_{\text{f rock 2}})[/latex]
• [latex]2.0 \text{ Ns West } + 0 \text{ Ns } = 0 \text{ Ns } + (4.0 \text{ kg})(\vec{v}_{\text{f rock 2}})[/latex]
• [latex](\vec{v}_{\text{f rock 2}}) = 2.0 \text{ Ns West } \div 4.0 \text{ kg or } 0.5 \text{ m/s West}[/latex]

During a tackle in a football game, two players of mass 90 kg and 70 kg were moving towards each other, and after collision both came to a full stop.  If the velocity of the larger player before collision was 7.0 m/s what was the velocity of the smaller player?

Solution

All we have for the direction in this problem is that both players were running towards each other.

Data:

• [latex]m_{90} = 90 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i 90}} = 7.0 \text{ m/s}[/latex]
• [latex]\vec{v}_{\text{f 90}} = 0 \text{ m/s}[/latex]

• [latex]m_{70} = 70 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i 70}} = \text{ Find}[/latex]
• [latex]\vec{v}_{\text{f 70}} = 0 \text{ m/s}[/latex]

Solution:

• [latex]\vec{p}_{\text{i 90}} + \vec{p}_{\text{i 70}} = \vec{p}_{\text{f 90}} + \vec{p}_{\text{f 70}}[/latex]
• [latex](m_{\text{90}})(\vec{v}_{\text{i 90}}) + (m_{\text{70}})(\vec{v}_{\text{i 70}}) = (m_{\text{90}})(\vec{v}_{\text{f 90}}) + (m_{\text{70}})(\vec{v}_{\text{f 70}})[/latex]
• [latex](90 \text{ kg})(7.0 \text{ m/s}) + (70 \text{ kg})(\vec{v}_{\text{i 70}}) = (90 \text{ kg})(0 \text{ m/s}) + (70 \text{ kg})(0 \text{ m/s})[/latex]
• [latex]630 \text{ Ns } + (70 \text{ kg})(\vec{v}_{\text{i 70}}) = 0 \text{ Ns } + 0 \text{ Ns}[/latex]
• [latex](\vec{v}_{\text{i 70}}) = - 630 \text{ Ns } \div 70 \text{ kg or } - 9.0 \text{ m/s}[/latex]

Note:  The [latex]- 9.0[/latex] m/s means that the smaller player was running in the opposite direction (towards) as the larger player.

If a 16 kg medicine ball is thrown horizontally to a stationary 65 kg skater on ice (consider this to be frictionless) and causes the skater who is now holding the ball to move off with a horizontal velocity of 0.50 m/s, what was the velocity the ball was thrown at?

Solution

Data:

• [latex]m_{16} = 16 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i 16}} = \text{ Find}[/latex]
• [latex]\vec{v}_{\text{f 16}} = 0.50 \text{ m/s}[/latex]

• [latex]m_{65} = 70 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{i 65}} = 0 \text{ m/s}[/latex]
• [latex]\vec{v}_{\text{f 65}} = 0.50 \text{ m/s}[/latex]

Solution:

• [latex]\vec{p}_{\text{i 16}} + \vec{p}_{\text{i 65}} = \vec{p}_{\text{f 16}} + \vec{p}_{\text{f 65}}[/latex]
• [latex](m_{\text{16}})(\vec{v}_{\text{i 16}}) + (m_{\text{65}})(\vec{v}_{\text{i 65}}) = (m_{\text{16}})(\vec{v}_{\text{f 16}}) + (m_{\text{65}})(\vec{v}_{\text{f 65}})[/latex]
• [latex](16 \text{ kg})(\vec{v}_{\text{i 16}}) + (65 \text{ kg})(0 \text{ m/s}) = (16 \text{ kg } + 65 \text{ kg})(0.5 \text{ m/s})[/latex]
• [latex](16 \text{ kg})(\vec{v}_{\text{i 16}}) + (0 \text{ Ns}) = (40.5 Ns)[/latex]
• [latex](\vec{v}_{\text{i 16}}) = 40.5 Ns \div 16 kg[/latex] ≈ 2.5 m/s horizontally

Note:  The ≈ + 2.5 m/s means that both the ball and the skater continue moving in the same horizontal direction as the medicine ball was thrown.

A 120 kg lineman is running at 8.5 m/s North and tackles a 85 kg halfback running at 9.6 m/s East.  At what velocity do these two move after the tackle occurs?

Solution

First, find the momenta of the two football players heading into the tackle.

The 85 kg Halfback:

• [latex]\vec{p}_{85} = m_{85} \vec{v}_{\text{i 85}}[/latex]
• [latex]p_{85} = (85 \text{ kg})(9.6 \text{ m/s East})[/latex] or 816 Ns East

The 120 kg Lineman:

• [latex]\vec{p}_{120} = m_{120} \vec{v}_{\text{ i 120}}[/latex]
• [latex]p_{120} = (120 \text{ kg})(8.5 \text{ m/s North})[/latex] or 1020 Ns North

The total momentum before the tackle is 816 Ns East + 1020 Ns North.

Now, in the absence of any other forces acting during this collision, the total momentum after the tackle for these two players must also be 816 Ns East + 1020 Ns North.

At this point, you will use trigonometry and the Pythagorean theorem to further analyze this collision.

Now: The total by using the momentum can be found using the Pythagorean Theorem.

• [latex](\vec{p}_{\text{total}})^2 = (816 \text{ Ns})^2 + (1020 \text{ Ns})^2[/latex]
• [latex](\vec{p}_{\text{total}})^2 = 665856 \text{ N}^2 \text{S}^2 + 1040400 \text{ N}^2 \text{S}^2[/latex]
• [latex](\vec{p}_{\text{total}})^2 = 1706256 \text{ N}^2 \text{S}^2[/latex]
• [latex]\vec{p}_{\text{total}} = 1306 \text{ Ns}[/latex]

The angle at  which these players move off is now found by using trigonometry, specifically the Tangent function:

• [latex]\tan ø = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}}[/latex]
• [latex]\tan ø = 1.245[/latex]
• [latex]ø = \tan ^{-1} 1.245[/latex]
• ø = 51.3°, specifically ≈ 51° North from East

The velocity at which these players move off can be found by using the equation for momentum:

• [latex]\vec{p} = m \vec{v}[/latex]
• [latex]1306 \text{ Ns} = (120 \text{ kg } + 85 \text{ kg}) \vec{v}_{\text{f (120 + 85)}}[/latex]
• [latex]\vec{v}_{\text{f (120 + 85)}} = 1306 \text{ Ns } \div 205 \text{ kg}[/latex]
• [latex]\vec{v}_{\text{f (120 + 85)}} = 64. \text{ m/s}[/latex] 51° North from East

Accident investigators are analyzing a car crash to see if excessive speed was involved. One of the cars had a mass of 1250 kg and was traveling at 40 km/h East when it was struck by a 1600 kg pickup truck traveling South that ran a red light. If the wreckage moved at 23.5° East from South, should the truck driver also be charged for driving above the 60 km/h speed limit?

Solution

First, draw a sketch of the two vehicles:

All that is known are the mass of the two colliding vehicles, original velocity of the car and the angle at which the wreck moved off.

Combining this data, we can calculate the momentum of the car, and by using trigonometry we can calculate the momentum of the truck.  The momentum of the truck allows us to estimate the initial velocity that it was traveling at.

First, the momentum of the car:

• [latex]\vec{p} = m \vec{v}[/latex]
• [latex]\vec{p} = (1250 \text{ kg})(11.1 \text{ m/s East})[/latex]
• [latex]\vec{p} = 13\;900 \text{ Ns East}[/latex]

Now, using the Tangent function:

• [latex]\tan 23.5^{\circ} = \dfrac{13\;900 \text{ Ns}}{\vec{p}_{\text{truck}}}[/latex]
• [latex]\vec{p}_{\text{truck}} = \dfrac{13\;900 \text{ Ns}}{\tan 23.5^{\circ}}[/latex]
• [latex]\vec{p}_{\text{truck}} = 32\;000 \text{ Ns South}[/latex]

Finally, the velocity of the truck is:

• [latex]\vec{p} = m \vec{v}[/latex]
• [latex]\vec{v} = (32\;000 \text{ Ns South}) \div (1600 \text{ kg})[/latex]
• [latex]\vec{v} = 20[/latex] m/s South or 72 km/h

The truck was speeding by 12 km/h and could also be charged with speeding.