Keep direction signs in momentum and square speeds only for kinetic energy.
Syllabus 4.3
Notes and key methods
Linear momentum is mass times signed velocity along the chosen line. Momenta in opposite directions have opposite signs. Total momentum is their algebraic sum, rather than the sum of the two momentum magnitudes.
Kinetic energy is one half mass times speed squared and is nonnegative. It has different units and different conservation conditions from momentum. A system with zero total momentum can still have positive kinetic energy because its particles may move in opposite directions.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Signed linear momentum
Take right as positive. At one instant, A has mass 5 kg and moves left at \(\displaystyle \frac{11m}{s}\); B has mass 4 kg and moves right at \(\displaystyle \frac{3m}{s}\).
Find A’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find B’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total kinetic energy in J.
Show the reasoning
(a)
Multiply mass by signed velocity.
A has velocity \(\displaystyle - \frac{11m}{s}\), giving momentum \(\displaystyle - \frac{55\mathrm{kg}m}{s}\).
Answer: \(-55\).
(b)
Use the common positive direction for the other particle.
The algebraic sum is \(\displaystyle - 55+ 12= - \frac{43\mathrm{kg}m}{s}\).
Answer: \(-43\).
(d)
Square each speed in its separate half-mass-times-speed-squared term.
Add the two nonnegative energies to obtain \(\displaystyle \frac{641}{2}\) J.
Answer: \(\displaystyle \frac{641}{2}\).
Example 2 · Signed linear momentum
Take right as positive. At one instant, A has mass 6 kg and moves left at \(\displaystyle \frac{4m}{s}\); B has mass 5 kg and moves right at \(\displaystyle \frac{4m}{s}\).
Find A’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find B’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total kinetic energy in J.
Show the reasoning
(a)
Multiply mass by signed velocity.
A has velocity \(\displaystyle - \frac{4m}{s}\), giving momentum \(\displaystyle - \frac{24\mathrm{kg}m}{s}\).
Answer: \(-24\).
(b)
Use the common positive direction for the other particle.
The algebraic sum is \(\displaystyle - 24+ 20= - \frac{4\mathrm{kg}m}{s}\).
Answer: \(-4\).
(d)
Square each speed in its separate half-mass-times-speed-squared term.
Add the two nonnegative energies to obtain \(88\) J.
Answer: \(88\).
Example 3 · Signed linear momentum
Take right as positive. At one instant, A has mass 7 kg and moves left at \(\displaystyle \frac{5m}{s}\); B has mass 6 kg and moves right at \(\displaystyle \frac{5m}{s}\).
Find A’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find B’s signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total signed momentum in \(\displaystyle \frac{\mathrm{kg}m}{s}\).
Find the total kinetic energy in J.
Show the reasoning
(a)
Multiply mass by signed velocity.
A has velocity \(\displaystyle - \frac{5m}{s}\), giving momentum \(\displaystyle - \frac{35\mathrm{kg}m}{s}\).
Answer: \(-35\).
(b)
Use the common positive direction for the other particle.