1d momentum
In the James Bond movie Diamonds are Forever the lead female character fires a machine gun while standing on the edge of an off-shore drilling rig. As she fires the gun she is driven back over the edge into the sea. The mass of a bullet is \( 10 g \) and its velocity is \( 750 \frac{m}{s} \). If her mass (including the gun) is \( 55 kg \), what recoil velocity does she acquire in response to a single shot from a stationary position?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
INFORMATION GIVEN
- \( m_{1} = 55kg \)
- \( v_{1i} = \text{?} \)
- \( m_{2} = 0.010kg \)
- \( v_{2i} = +750 \frac{m}{s} \)
USEFUL FORMULA(S)
- \( p = mv \)
- \( m_{1} v_{1f} + m_{2} v_{2f} = m_{1} v_{1i} + m_{2} v_{2i} \)
PROCESS
- Transform the conservation equation:
- \( 0 = m_{1} v_{1i} + m_{2} v_{2i} \)
- \( -m_{1} v_{1i} = m_{2} v_{2i} \)
- \( \Rightarrow v_{1i} = \frac{ -m_{2} v_{2i} }{ m_{1} } \)
- \( v_{1i} = \frac{ -m_{2} v_{2i} }{ m_{1} } \)
- \( v_{1i} = \frac{ - ( 0.010kg ) ( 750\frac{m}{s} ) }{ 55kg } \)
- \( \Rightarrow v_{1i} = -0.14 \frac{m}{s} \)
- She acquired \( -0.14 \frac{m}{s} \) in recoil velocity.
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case, we're dealing with momentum. We grabbed the formulas for momentum and conservation of momentum.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We eliminated the 'final' half of the conservation of momentum formula, as we didn't need it, then rearranged to isolate \( v_{1i} \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
Energy and work
Fred is moving into an apartment at the beginning of the school year. Fred weighs \( 685 N \) and his belongings weigh \( 915 N \). How much work does the elevator do in lifting Fred and his belongings up five stories (\( 15.2 m \))? How much work does the elevator do on Fred on the downward trip?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
INFORMATION GIVEN
- \( F_{1} = 685N \)
- \( F_{2} = 915N \)
- \( d = 15.2m \)
- \( W_{up} = \text{?} \)
- \( W_{down} = \text{?} \)
USEFUL FORMULA(S)
- \( F_{net} = F_{1} + F_{2} + \cdots + F_{n} \)
- \( W = F \Delta d \)
PROCESS
- \( F_{net} = 685N + 915N = 1600N \)
- On the way up:
- \( W_{up} = F_{net} \Delta d \)
- \( W_{up} = (1600N) (15.2m) \)
- \( \Rightarrow W_{up} = +2.43 \times 10^{4} J \)
- On the way down:
- \( W_{down} = F_{1} \Delta d \)
- \( W_{down} = (685N) (-15.2m) \)
- \( \Rightarrow W_{down} = -1.04 \times 10^{4} J \)
- The elevator did \( +2.43 \times 10^{4} J \) of work lifting Fred and his belongings up five stories, and did \( -1.04 \times 10^{4} J \) of work on the downward trip.
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case, we're dealing with work and energy. We grabbed the formulas for work and net force.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We didn't have to do that this time.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
- We first solved for net force, then went on to find the work done for both directions.
Conservation of energy
![Picture](/uploads/6/1/5/8/61582483/fdbpbaboheicbooj_orig.png)
A water skier lets go of the tow rope upon leaving the end of a jump ramp at a speed of \( 14.0 \frac{m}{s} \). As the drawing indicates, the skier has a speed of \( 13.0 \frac{m}{s} \) at the highest point of the jump. Ignoring air resistance, determine the skier’s height H above the top of the ramp at the highest point.
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( v_{i} = 14.0 \frac{m}{s} \)
- \( v_{f} = 13.0 \frac{m}{s} \)
- \( g = 9.81 \frac{m}{s^{2}} \)
- \( E_{gi} = 0 \)
- \( h = \text{?} \)
USEFUL FORMULA(S)
- \( E_{gi} + E_{Ki} = E_{gf} + E_{Kf} \)
PROCESS
- Transform the main equation:
- \( E_{gi} + E_{Ki} = E_{gf} + E_{Kf} \)
- \( E_{Ki} = E_{Kf} + E_{gf} \)
- \( E_{gf} = E_{Ki} - E_{Kf} \)
- \( mgh = ( \frac{1}{2} mv^{2}_{i} ) - ( \frac{1}{2} mv^{2}_{f} ) \)
- \( mgh = \frac{ mv^{2}_{i} }{2} - \frac{ mv^{2}_{f} }{2} \)
- \( mgh = \frac{ mv^{2}_{i} - mv^{2}_{f} }{2} \)
- \( h = \frac{ v^{2}_{i} - v^{2}_{f} }{ 2g } \)
- \( h = \frac{ v^{2}_{i} - v^{2}_{f} }{ 2g } \)
- \( h = \frac{ ( 14.0 \frac{m}{s} )^{2} - ( 13.0 \frac{m}{s} )^{2} }{ 2 ( 9.81 \frac{ m }{ s^{2} } ) } \)
- \( \Rightarrow h = 1.38m \)
- The water skier reached a maximum height of \( 1.38m \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case, we're dealing with the law of conservation of energy. We grabbed the formula for, you guessed it, conservation of energy.
- Break down the useful formula(s) into the components of information available to you.
- This was quite the process. We had to break everything down to its most basic pieces, as well as think cleverly with regards to gravitational potential energy.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We rearranged to isolate \( h \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
power
What is the power output of a machine which applies a force of \( 2.50 \times 10^{4} N \) for \( 12.0 s \) in pulling a block through \( 60.0 m \)?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
INFORMATION GIVEN
- \( F = 2.50 \times 10^{4} N \)
- \( d = 60.0m \)
- \( t = 12.0s \)
- \( P = \text{?} \)
USEFUL FORMULA(S)
- \( P = \frac{ W }{ t } \)
PROCESS
- Transform the main equation:
- \( P = \frac{ W }{ t } \)
- \( P = \frac{ F \Delta d }{ t } \)
- \( P = \frac{ F \Delta d }{ t } \)
- \( P = \frac{ (2.5 \times 10^{4} N )( 60.0m ) }{ 12.0s } \)
- \( \Rightarrow P = 125000 W \rightarrow 125kW \)
- The machine gave an output of \( 125kW \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case, we're dealing with power. We grabbed the formula for, big surprise, power.
- Break down the useful formula(s) into the components of information available to you.
- We only had to break the power equation down a little bit, fortunately.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We didn't have to rearrange any equations for this question.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
efficiency
A lift is able to carry a load of \( 1000 kg \) at a velocity of \( 1.5 \frac{m}{s} \). Given that the motor driving the lift has an input power of \( 20 kW \), calculate the efficiency of the lift.
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
INFORMATION GIVEN
- \( m = 1000kg \)
- \( v = 1.5 \frac{m}{s} \)
- \( g = 9.81 \frac{m}{ s^{2} } \)
- \( P_{i} = 2.0 \times 10^{4} W \)
- \( P_{f} = \text{?} \)
- \( F = \text{?} \)
- \( \eta = \text{?} \)
USEFUL FORMULA(S)
- \( \eta = \frac{ P_{f} }{ P_{i} } \cdot 100 \)
- \( P = \frac{ W }{ t } \)
- \( W = F \Delta d \)
- \( v = \frac{ \Delta d }{ t } \)
- \( F = mg \)
PROCESS
- Transform power equation:
- \( P = \frac{ W }{ t } \)
- \( P = \frac{ F \Delta d }{ t } \)
- \( P = F \cdot \frac{ \Delta d }{ t } \)
- \( P = F \cdot v \)
- \( F = mg \)
- \( F = ( 1000kg )( 9.81 \frac{m}{ s^{2} } ) \)
- \( \Rightarrow F = 9810N \)
- \( P_{f} = F \cdot v \)
- \( P_{f} = ( 9810N )( 1.5 \frac{m}{s} ) \)
- \( \Rightarrow P_{f} = 1.5 \times 10^{4} \)
- \( \eta = \frac{ P_{f} }{ P_{i} } \cdot 100 \)
- \( \eta = \frac{ 1.5 \times 10^{4} W }{ 2.0 \times 10^{4} W } \cdot 100 \)
- \( \Rightarrow \eta = 75 \% \)
- The lift is running with \( 75 \% \) efficiency.
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case, we're dealing with efficiency. We grabbed the formulas for efficiency, power, work, force, and velocity.
- Break down the useful formula(s) into the components of information available to you.
- This was quite the process. We had to break the efficiency formula waaay down to fit the few pieces of information we had.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We didn't have to rearrange, just break down.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
- We first solved for force, then output work, then lastly the efficiency.
2d momentum
A \( 100 kg \) mass explodes into three parts. The first part travels away at \( 50 \frac{m}{s} \) straight NORTH and has a mass of \( 20 kg \). The second part travels away at \( 35 \frac{m}{s} \) straight WEST and has a mass of \( 50 kg \). What is the resultant velocity of the third part?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
THE CHALLENGE
This was a very complicated question, and of a type I hadn't encountered before. It called for two different areas of math (momentum and trigonometry) that had more in common than I realized, and made me think more carefully. I recognized that the known pieces had very specific paths, but it took me a while to see how I could use that information.
THE SOLUTION
When I realized how I needed to utilize the Pythagorean Theorem and trigonometry, things started to fall into place. I needed to visualize the paths of the pieces as the legs of a triangle, with specific lengths corresponding to their speeds. Using this method, I was able to calculate the momentum of the third piece, which in turn lead to the use of trigonometric ratios to find the direction of motion for the third piece.
INFORMATION GIVEN
- \( m_{100} = 100 kg \)
- \( m_{50} = 50 kg \)
- \( m_{20} = 20 kg \)
- \( m_{30} = \text{?} \)
- \( v_{100} = 0 \frac{m}{s} \)
- \( v_{50} = 35 \frac{m}{s} \ \text{West} \)
- \( v_{20} = 50 \frac{m}{s} \ \text{North} \)
- \( v_{30} = \text{?} \)
USEFUL FORMULA(S)
- \( a^{2} + b^{2} = c^{2} \)
- \( p = mv \)
- \( tan \theta = \frac{opp}{adj} \)
PROCESS
- \( m_{30} = m_{100} - m_{50} - m_{20} \)
- \( m_{30} = 100kg - 50kg - 20kg \)
- \( \Rightarrow m_{30} = 30 kg \)
- \( p_{100} = m_{100} \cdot v_{100} \)
- \( p_{100} = ( 100kg )( 0 \frac{m}{s} ) \)
- \( \Rightarrow p_{100} = 0 \frac{kg \cdot m}{s} \)
- \( p_{50} = m_{50} \cdot v_{50} \)
- \( p_{50} = ( 50kg )( 35 \frac{m}{s} ) \)
- \( \Rightarrow p_{50} = 1750 \frac{kg \cdot m}{s} \ \text{West} \)
- \( p_{20} = m_{20} \cdot v_{50} \)
- \( p_{20} = ( 20kg )( 50 \frac{m}{s} ) \)
- \( \Rightarrow p_{20} = 1000 \frac{kg \cdot m}{s} \ \text{North} \)
- \( p_{30} = \text{?} \)
- Transform the Pythagorean Theorem:
- \( c^{2} = a^{2} + b^{2} \)
- \( \Rightarrow c = \sqrt{ a^{2} + b^{2} } \)
- \( c = \sqrt{ a^{2} + b^{2} } \)
- \( p_{30} = \sqrt{ p_{20}^{2} + p_{50}^{2} } \)
- \( p_{30} = \sqrt{ 1000^{2} + 1750^{2} } \)
- \( \Rightarrow p_{30} \approx 2.0 \times 10^{3} \frac{kg \cdot m}{s}\)
- \( v_{30} = \text{?} \)
- Transform the momentum equation:
- \( p = mv \)
- \( \Rightarrow v = \frac{p}{m} \)
- \( v = \frac{p}{m} \)
- \( v_{30} = \frac{ p_{30} }{ m_{30} } \)
- \( v_{30} = \frac{ 2.0 \times 10^{3} \frac{kg \cdot m}{s} }{ 30kg } \)
- \( \Rightarrow v_{30} \approx 66.7 \frac{m}{s} \)
- Still need direction.
- \( tan \theta = \frac{opp}{adj} \)
- \( \theta = tan^{-1} ( \frac{opp}{adj} ) \)
- \( \theta = tan^{-1} ( \frac{ p_{20} }{ p_{50} } ) \)
- \( \theta = tan^{-1} ( \frac{ 1000 }{ 1750 } ) \)
- \( \Rightarrow \theta \approx 30^{\circ} \)
- The third part of the mass was launched at roughly \( 66.7 \frac{m}{s} \), \( 30^{\circ} \) South of East.
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with momentum. We grabbed the formulas for momentum, as well as the Pythagorean Theorem and the trigonometric ratio for tangent.
- Break down the useful formula(s) into the components of information available to you.
- We didn't have to break down our formulas too much, but we did have to substitute values quite frequently.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the momentum formula, as well as the tangent ratio.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
- We first solved for the mass of the third piece, then calculated the momentum of the other, known parts. We used the Pythagorean Theorem to calculate the momentum of our third piece, and rearranged the momentum formula to solve for the velocity of our third piece. Lastly, it was a simple matter of some trigonometry to find the direction of motion for the third piece.
THE CHALLENGE
This was a very complicated question, and of a type I hadn't encountered before. It called for two different areas of math (momentum and trigonometry) that had more in common than I realized, and made me think more carefully. I recognized that the known pieces had very specific paths, but it took me a while to see how I could use that information.
THE SOLUTION
When I realized how I needed to utilize the Pythagorean Theorem and trigonometry, things started to fall into place. I needed to visualize the paths of the pieces as the legs of a triangle, with specific lengths corresponding to their speeds. Using this method, I was able to calculate the momentum of the third piece, which in turn lead to the use of trigonometric ratios to find the direction of motion for the third piece.