(c) A child drops an egg from a height of $10 \mathrm{~cm}$ and the egg lands on the floor.
Explain why the egg is less likely to break if the floor is covered with a thick carpet than if the floor were covered in hard tiles.
2 This question is about momentum.
(a) Which of these is the correct unit for momentum?
(1)
A $\mathrm{kg} / \mathrm{m} / \mathrm{s}$
B $\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}$
C $\mathrm{kgm} / \mathrm{s}^{2}$
D $\mathrm{kgm} / \mathrm{s}$
(b) The diagram shows an object before and after an explosion.
The object breaks into two parts, $P$ and $Q$.
The parts move away from each other in opposite directions.
Before
(1)
(ii) Calculate the magnitude of the velocity of part $\mathrm{P}$ after the explosion.
(3)
velocity $=$
$\mathrm{m} / \mathrm{s}$

Question

Accepted Answer

Sure, let's address the questions based on the provided images and text. ### Question 1: Egg Drop Scenario ∻Answer∻ ⚹ The egg is less likely to break if the floor is covered with a thick carpet. ⚹ ∻Solution∻ ‖ a ⋮ Impact Force: When the egg lands on a thick carpet, the carpet increases the time over which the egg's momentum is brought to zero. According to the impulse-momentum theorem, $F \Delta t = \Delta p$, where $F$ is the force, $\Delta t$ is the time interval, and $\Delta p$ is the change in momentum. ‖ ‖ b ⋮ Reduced Force: By increasing the time interval $\Delta t$, the force $F$ exerted on the egg is reduced, making it less likely to break. ‖ ∻Key Concept∻ ⚹ Impulse-Momentum Theorem ⚹ ∻Explanation∻ ⚹ The thick carpet increases the time over which the egg's momentum changes, reducing the impact force and thus the likelihood of the egg breaking. ⚹ ### Question 2: Momentum Calculation ∻Answer∻ ⚹ The correct unit for momentum is $\mathrm{kgm} / \mathrm{s}$. The magnitude of the velocity of part P after the explosion is $2.42 \, \mathrm{m/s}$. ⚹ ∻Solution∻ ‖ a ⋮ Unit of Momentum: The correct unit for momentum is $\mathrm{kgm} / \mathrm{s}$ (Option D). ‖ ‖ b ⋮ Conservation of Momentum: According to the principle of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. ‖ ‖ c ⋮ Initial Momentum: Before the explosion, the object is stationary, so the initial momentum is $0$. ‖ ‖ d ⋮ Final Momentum: After the explosion, the momentum of part P and part Q must sum to zero. Let $v$ be the velocity of part P. The equation is: $$ m_P v_P + m_Q v_Q = 0 $$ $$ 2.9 \, \mathrm{kg} \cdot v + 1.9 \, \mathrm{kg} \cdot 3.7 \, \mathrm{m/s} = 0 $$ ‖ e ⋮ Solving for $v$: $$ 2.9v = -1.9 \cdot 3.7 $$ $$ v = \frac{-1.9 \cdot 3.7}{2.9} $$ $$ v = -2.42 \, \mathrm{m/s} $$ The magnitude of the velocity is $2.42 \, \mathrm{m/s}$. ‖ ∻Key Concept∻ ⚹ Conservation of Momentum ⚹ ∻Explanation∻ ⚹ The total momentum before and after the explosion must be equal. By setting up the equation and solving for the unknown velocity, we find the magnitude of the velocity of part P. ⚹◊How does the principle of conservation of momentum apply to the scenario described in the image?⍭ Generate me a similar question◊‖

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