Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — kinematics by calculus
A particle has v=3t2−12t+9 m s−1 and s=0 at t=0. Find when it is at rest and its displacement from t=1 to t=3.
- 1At rest: 3t2−12t+9=0Set the velocity, not the displacement, to zero.
- 23(t−1)(t−3)=0, so t=1 s and t=3 s
- 3s=∫vdt=t3−6t2+9t+c, and s(0)=0 gives c=0The initial condition fixes the constant.
- 4s(1)=1−6+9=4 and s(3)=27−54+27=0
- 5Displacement =0−4=−4 mNegative: the particle has moved back. The DISTANCE travelled is 4 m.
Example 2 — connected particles over a pulley
Masses of 5 kg and 3 kg hang from a light inextensible string over a smooth pulley. Find the acceleration and the tension. Take g=9.81.
- 1For the 5 kg mass (moving down): 5g−T=5aTake the direction of motion as positive for each mass.
- 2For the 3 kg mass (moving up): T−3g=3aSame T and same a — that is what light and inextensible mean.
- 3Adding: 2g=8aAdding eliminates T.
- 4a=82(9.81)=2.45 m s−2
- 5T=3(g+a)=3(12.26)=36.8 NBetween 3g=29.4 N and 5g=49.1 N ✓
Example 3 — a collision with coalescence
A 2 kg body moving at 5 m s−1 strikes a stationary 3 kg body and they move off together. Find the common velocity and the impulse on the 3 kg body.
- 1Momentum before =2(5)+3(0)=10 kg m s−1Momentum is conserved even though energy is not.
- 2Momentum after =(2+3)v=5vThey coalesce, so a single mass moves.
- 35v=10, so v=2 m s−1Less than 5 — the mass has increased.
- 4Impulse on the 3 kg body =mv−mu=3(2)−0=6 N s
- 5On the 2 kg body: 2(2)−2(5)=−6 N sEqual and opposite ✓ — Newton's third law.