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iGCSE Physics (0625) 1.6 Momentum Paper 4 -Exam Style Questions- New Syllabus

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Question

Fig. 3.1 shows two toy trains, A and B, on a track.
The mass of train A is $0.45\text{ kg}$ and the mass of train B is $0.21\text{ kg}$. The trains do not have motors.
Train A travels towards train B and they collide.
Immediately before the collision:
  • the velocity of train A is $0.34\text{ m/s}$ to the right
  • the velocity of train B is $0.12\text{ m/s}$ to the left
The trains stick together when they collide.
(a) Define momentum.
(b) Calculate the velocity of the trains immediately after the collision.
(c) A different train is travelling with a momentum of $0.26\text{ kg m/s}$. The train slows down and stops after $2.1\text{ s}$. Calculate the average force acting on the train during this time.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic $1.6$ — Momentum (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)

▶️ Answer/Explanation

(a)
For the correct answer:
$\text{mass} \times \text{velocity}$

Momentum is a vector quantity defined as the product of an object’s mass and its velocity. In mathematical terms, it is expressed by the equation $p = mv$.

(b)
For the correct answer:
$0.19\text{ m/s}$ to the right

Using the conservation of momentum: $\text{Total initial momentum} = (0.45 \times 0.34) + (0.21 \times -0.12) = 0.153 – 0.0252 = 0.1278\text{ kg m/s}$. Since they stick together, the final mass is $0.66\text{ kg}$. $v = \frac{0.1278}{0.66} \approx 0.19\text{ m/s}$.

(c)
For the correct answer:
$0.12\text{ N}$ (or $-0.12\text{ N}$)

Resultant force is the rate of change of momentum: $F = \frac{\Delta p}{\Delta t} = \frac{0 – 0.26}{2.1} \approx -0.12\text{ N}$. The magnitude of the average force is $0.12\text{ N}$.

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