This set practises the five skills in forces and momentum: drawing forces, finding a resultant, using F = ma, conserving momentum and reading balanced forces. All situations and numbers are invented for practice. Where weight is needed, take g = 10 N/kg.
Work on paper, write units, and give directions for vectors. Questions run from easy to harder. Open the answer only after you have written your own.
Questions
Q1. A trolley is pushed to the right with 25 N and opposed by 15 N of friction to the left. Find the resultant force.
Show answer
Right is positive: 25 − 15 = 10 N to the right.
Q2. A resultant force of 15 N acts on a 6.0 kg trolley. Find its acceleration.
Show answer
a = F/m = 15 ÷ 6.0 = 2.5 m/s² in the direction of the force. Check: 6.0 × 2.5 = 15 N.
Q3. A cyclist rides along a level road at a steady speed. List the forces on the cyclist and bicycle, and say how the horizontal forces compare.
Show answer
Weight (down), normal force (up), driving force (forward) and air resistance plus friction (backward). Steady speed means zero acceleration, so the forward and backward forces are equal and the resultant is zero.
Q4. An object moves in a straight line at 4.0 m/s. A student says, “It must have a forward resultant force”. Is this correct?
Show answer
No. Steady velocity means zero acceleration. So the resultant force is zero. A force is needed to change the velocity, not to maintain it.
Q5. Calculate the momentum of a 0.45 kg football moving at 20 m/s.
Show answer
p = mv = 0.45 × 20 = 9.0 kg m/s in the direction of motion.
Q6. A resultant force of 35 N accelerates an object at 0.50 m/s². Find its mass.
Show answer
m = F/a = 35 ÷ 0.50 = 70 kg. Check: 70 × 0.50 = 35 N.
Q7. A lift cable pulls upwards with 6500 N. The lift has a weight of 6000 N. Find the resultant force, the mass of the lift and its acceleration.
Show answer
Up is positive: resultant = 6500 − 6000 = 500 N upwards. Mass = weight ÷ g = 6000 ÷ 10 = 600 kg. a = F/m = 500 ÷ 600 = 0.833 = 0.83 m/s² upwards. Check: 600 × 0.833 = 500 N.
Q8. A 1.5 kg trolley moving at 4.0 m/s hits a stationary 0.50 kg trolley and they stick together. Find their common velocity.
Show answer
Before: 1.5 × 4.0 = 6.0 kg m/s. After: (1.5 + 0.50) × v = 2.0v. So v = 6.0 ÷ 2.0 = 3.0 m/s in the original direction. Check: 2.0 × 3.0 = 6.0.
Q9. A parachutist of mass 70 kg falls at steady speed before opening the parachute. Then the parachute opens and the air resistance suddenly becomes 1500 N. Find the air resistance before opening, and the resultant force and acceleration just after.
Show answer
Weight = 70 × 10 = 700 N. Steady speed means balanced forces, so the air resistance before opening is 700 N upwards. Just after: up is positive, 1500 − 700 = 800 N upwards. a = 800 ÷ 70 = 11.4 = 11 m/s² upwards (the parachutist slows down). Check: 70 × 11.43 ≈ 800 N.
Q10. Cart A, 2.0 kg at 5.0 m/s to the east, collides head-on with cart B, 3.0 kg at 2.0 m/s to the west. They stick together. Find the common velocity.
Show answer
East is positive. Before: 2.0 × 5.0 + 3.0 × (−2.0) = 10 − 6.0 = 4.0 kg m/s. After: 5.0 × v. So v = 4.0 ÷ 5.0 = 0.80 m/s to the east. Check: 5.0 × 0.80 = 4.0.
Q11. A 60 kg skater pushes a 40 kg skater away while both are at rest on smooth ice. The 40 kg skater moves at 1.5 m/s. Find the velocity of the 60 kg skater.
Show answer
Total momentum before is zero. The 40 kg skater has 40 × 1.5 = 60 kg m/s one way. So the 60 kg skater has 60 kg m/s the other way: v = 60 ÷ 60 = 1.0 m/s in the opposite direction.
Q12. Two forces of 6.0 N east and 8.0 N north act on an object. Find the size of the resultant force. (Check your syllabus year for whether direction is needed.)
Show answer
√(6.0² + 8.0²) = √(36 + 64) = √100 = 10 N. The direction is tan⁻¹(8.0 ÷ 6.0) = 53° north of east (to the nearest degree).
If you got these wrong
Match each slip with the lesson that fixes it.
| What went wrong | Go to |
|---|---|
| Missed an arrow, added a “motion” arrow or drew a force on the wrong object (Q3) | Draw a force diagram |
| Added opposing forces, or forgot direction (Q1, Q7, Q12) | Calculate a resultant force |
| Used one force instead of the resultant, or left grams unconverted (Q2, Q6, Q7, Q9) | Relate acceleration to resultant force |
| Forgot signs or treated a collision as one-directional (Q5, Q8, Q10, Q11) | Apply momentum conservation |
| Said a moving object needs a force, or that zero resultant means no forces (Q3, Q4, Q9) | Balanced forces versus no forces |
What should you do next?
Record each miss in the mistake log and retest queue, then redo the question in two or three days with new numbers. Return to the module overview if the order of ideas feels unclear.
If one error type keeps coming back after a retest, a teacher in online one-to-one Physics tuition can look at your working and address it directly.