Momentum

Calculate — 5 marks

A railway coupling system is being tested. An empty freight wagon of mass 8000 kg is moving at 3 m/s along a straight horizontal track. It collides and couples with a stationary loaded freight wagon of mass 12000 kg. The two wagons then move together after the collision.

(a) Calculate the momentum of the empty freight wagon before the collision. [1 mark]
(b) Using the principle of conservation of momentum, calculate the velocity of the two coupled wagons immediately after the collision. [3 marks]
(c) Calculate the loss in kinetic energy during the collision. (You may use: kinetic energy = ½mv²) [1 mark]

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(a) p = mv = 8000 × 3 = 24000 kg m/s (or 2.4 × 10⁴ kg m/s) — 1 mark
(b) Total momentum before = momentum after — 1 mark
(b) (8000 × 3) + (12000 × 0) = (8000 + 12000) × v — 1 mark
(b) 24000 = 20000v; v = 1.2 m/s — 1 mark
(c) KE before = ½ × 8000 × 3² = 36000 J; KE after = ½ × 20000 × 1.2² = 14400 J; Loss = 36000 − 14400 = 21600 J (or 2.16 × 10⁴ J) — 1 mark

State — 4 marks

A delivery driver is loading packages into a van. Package A (mass 5 kg) is moving at 2 m/s and collides with Package B (mass 3 kg) which is initially at rest on a frictionless conveyor belt.

(a) State what is meant by momentum. [1 mark]
(b) State the momentum of Package A before the collision. [1 mark]
(c) State the principle of conservation of momentum and explain why it applies to this collision between the two packages. [2 marks]

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(a) Momentum is the product of mass and velocity / p = mv
(b) 10 kg m/s (or 10 N s)
(c) In a closed system / when no external forces act, total momentum before = total momentum after (or momentum is conserved)
(c) The collision occurs on a frictionless surface / there are no external forces acting on the system / friction is negligible

Show — 2 marks

A skateboard rider of mass 60 kg is moving at a constant velocity of 5 m/s along a flat street. The rider then pushes off the ground, accelerating to 8 m/s in 2 seconds.

Show that the momentum of the skateboard rider increases by 180 kg m/s during the acceleration. [2 marks]

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Calculates initial momentum: p = mv = 60 × 5 = 300 kg m/s (1 mark)
Calculates final momentum: p = mv = 60 × 8 = 480 kg m/s and finds change: 480 − 300 = 180 kg m/s (1 mark)

Explain — 5 marks

A railway shunting yard uses coupling systems to join freight cars. Two identical freight cars, each with a mass of 2000 kg, are moving on parallel tracks. Car A moves at 3 m/s to the right, while Car B moves at 1 m/s to the right on an adjacent track. The cars are brought together and couple automatically, moving as a single combined unit after collision.

(a) Explain why momentum is described as a vector quantity, using the scenario of the two freight cars. [2 marks]
(b) Calculate the velocity of the coupled cars immediately after they join together. Explain how the principle of conservation of momentum applies to this collision, including reference to the direction of motion. [3 marks]

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(a) Momentum has both magnitude and direction (1 mark)
(a) Direction matters because both cars move in the same direction but their combined momentum depends on their individual momenta as vectors (or equivalent explanation showing direction is essential) (1 mark)
(b) Correct calculation: total momentum before = (2000 × 3) + (2000 × 1) = 8000 kg m/s; total mass after = 4000 kg; final velocity = 8000 ÷ 4000 = 2 m/s (1 mark)
(b) Momentum is conserved because no external forces act on the system during the collision (1 mark)
(b) The final velocity is to the right because the total momentum before collision acts in that direction (1 mark)

Calculate — 5 marks

State — 4 marks

Show — 2 marks

Explain — 5 marks

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