A ball of mass 0.5 kg moves in a vertical circle of radius 10 m on the end of a string. At the lowest point its speed is 10 m s−1.
What is the tension in the string at the lowest point?
- A 0.095 N
- B 4.9 N
- C 5.0 N
- D 9.9 N
Show answer
Answer: D
Free A.2 Forces & momentum questions for IB Physics (SL & HL), part of A Space, time & motion. Each question has the answer and the marking points an examiner would look for.
Generate a full A.2 Forces & Momentum worksheet →All A Space, time & motion questionsA ball of mass 0.5 kg moves in a vertical circle of radius 10 m on the end of a string. At the lowest point its speed is 10 m s−1.
What is the tension in the string at the lowest point?
Answer: D
A skydiver reaches terminal speed and then opens her parachute. What happens to the drag force and the speed immediately after the parachute opens?
| drag force | speed | |
|---|---|---|
| A | stays the same | decreases |
| B | increases | increases |
| C | increases | decreases |
| D | decreases | decreases |
Answer: C
A steel ball of radius 1.5 mm and density 7800 kg m−3 is released from rest in a tall column of glycerol of density 1260 kg m−3 and viscosity 1.4 Pa s.
Show that the weight of the ball is about 1 × 10−3 N.
Calculate the initial acceleration of the ball.
Explain why the ball reaches a constant velocity.
Determine the terminal speed of the ball.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.08 × 10−3 N V = 4/3 πr3 = 1.41 × 10−8 m3 ✓ W = ρVg = 1.08 × 10−3 N ✓ | 2 |
| (b) | 8.2 m s−2 buoyancy = 1.75 × 10−4 N ✓ a = (W − B)/m = 8.2 m s−2 ✓ | 2 |
| (c) | drag increases with speed until forces balance viscous drag increases as the speed increases ✓ when drag + buoyancy = weight the resultant force is zero ✓ | 2 |
| (d) | 0.023 m s−1 6πηrv = W − B ✓ v = 2r2g(ρs − ρf)/9η ✓ = 0.023 m s−1 ✓ | 3 |
Glider A, of mass 0.3 kg, moves at 0.8 m s−1 along a horizontal air track towards a stationary glider B of mass 0.6 kg. After the collision A rebounds at 0.1 m s−1.
State the condition for the total momentum of the two gliders to be conserved.
Calculate the velocity of B after the collision.
Determine whether the collision is elastic.
The force on B during the collision varies with time as a triangle, lasting 0.08 s. Determine the maximum force on B.
Explain, with reference to Newton’s third law, why the change of momentum of A is equal and opposite to that of B.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | no external force no resultant external force acts on the system ✓ | 1 |
| (b) | 0.450 m s−1 0.3(0.8) = 0.3(-0.1) + 0.6v ✓ v = 0.450 m s−1 ✓ | 2 |
| (c) | inelastic initial KE = 0.0960 J ✓ final KE = 0.0623 J ✓ kinetic energy is not conserved, so inelastic ✓ | 3 |
| (d) | 6.75 N impulse = Δp = 0.270 N s = area = ½Fmaxt ✓ Fmax = 6.75 N ✓ | 2 |
| (e) | equal and opposite forces for same time the force of A on B is equal and opposite to the force of B on A ✓ they act for the same time, so the impulses are equal and opposite ✓ | 2 |
A car of mass 1200 kg travelling at 25 m s−1 brakes and stops in a distance of 40 m.
Calculate the deceleration of the car.
Calculate the average braking force.
Determine the minimum coefficient of friction between the tyres and the road.
Explain, with reference to momentum, why a seat belt that stretches reduces injury in a collision.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.8 m s−2 v2 = u2 + 2as ✓ a = 625/80 = 7.81 m s−2 ✓ | 2 |
| (b) | 9.4 kN F = 1200 × 7.81 = 9.4 × 103 N ✓ | 1 |
| (c) | 0.80 μmg = ma ✓ μ = 7.81/9.8 = 0.80 ✓ | 2 |
| (d) | longer time, smaller force the change of momentum of the passenger is the same ✓ it occurs over a longer time so the force is smaller ✓ | 2 |
A box of mass 12 kg is pulled along a horizontal floor by a rope at 30° above the horizontal. The tension is 60 N and the box accelerates at 1.5 m s−2.
Calculate the horizontal component of the tension.
Calculate the frictional force on the box.
Calculate the normal reaction on the box.
Hence determine the coefficient of dynamic friction.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 52 N 60 cos 30° = 52 N ✓ | 1 |
| (b) | 34 N Fnet = ma = 18 N ✓ friction = 52 − 18 = 34 N ✓ | 2 |
| (c) | 88 N N = mg − 60 sin 30° ✓ 118 − 30 = 88 N ✓ | 2 |
| (d) | 0.39 μ = F/N ✓ 0.39 ✓ | 2 |
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