Which of the following best describes the relationship between momentum and time for an object in free fall, starting from rest?

Here are the solutions to the questions:

Question 1.1

• Concept: Newton's Second Law of Motion states that the net force acting on an object is equal to the rate of change of its momentum.
• Formula: $F = \frac{\Delta p}{\Delta t}$
• Explanation: This formula directly defines force as the rate of change of momentum.

The correct option is D.
$\boxed{\text{D Force}}$

Question 1.2

• Concept: For an object in free fall (ignoring air friction), the acceleration due to gravity ($g$) is constant. Momentum ($p$) is the product of mass ($m$) and velocity ($v$).
• Step 1: Determine velocity at time $t$. Since the object is dropped, its initial velocity $u = 0$. Using the kinematic equation $v = u + at$, where $a = g$: $$v_t = 0 + gt = gt$$
• Step 2: Express momentum at time $t$. $$p = mv_t = mgt$$
• Step 3: Determine velocity at time $2t$. $$v_{2t} = 0 + g(2t) = 2gt$$
• Step 4: Express momentum at time $2t$. Let the momentum at time $2t$ be $p'$. $$p' = mv_{2t} = m(2gt)$$ $$p' = 2(mgt)$$
• Step 5: Relate $p'$ to $p$. Since $p = mgt$, we can substitute this into the expression for $p'$: $$p' = 2p$$

The correct option is C.
$\boxed{\text{C } 2p}$

Question 1.3

• Concept: According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. If the boy exerts a force on the box, the box exerts an equal and opposite force on the boy.
• Explanation: The boy pushes the box with a force of 250 N. Therefore, the box exerts a force of 250 N back on the boy. The frictional force of 50 N acts on the box, opposing its motion, but it is not part of the action-reaction pair between the boy and the box.

The correct option is C.
$\boxed{\text{C } 250 \text{ N}}$

Question 1.4

• Concept: Positional isomers are compounds that have the same molecular formula and the same carbon skeleton, but differ in the position of a functional group, substituent, or multiple bond.
• Step 1: Determine the molecular formula and identify the carbon skeleton for each compound.
  • I: $\text{CH} \equiv \text{C} - \text{CH}_2 - \text{CH}_3$
    • Molecular formula: $\text{C}_4\text{H}_6$
    • Name: But-1-yne (4-carbon chain, triple bond at position 1)
  • II: $\text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_2 - \text{CH}_3$
    • Molecular formula: $\text{C}_5\text{H}_8$
    • Name: Pent-2-yne (5-carbon chain, triple bond at position 2)
  • III: $\text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_3$
    • Molecular formula: $\text{C}_4\text{H}_6$
    • Name: But-2-yne (4-carbon chain, triple bond at position 2)
  • IV: $\text{CH}_3 - \text{C} \equiv \text{CH}$
    • Molecular formula: $\text{C}_3\text{H}_4$
    • Name: Propyne (3-carbon chain, triple bond at position 1)
• Step 2: Identify compounds with the same molecular formula and carbon skeleton but different positions of the multiple bond.
  • Compounds I and III both have the molecular formula $\text{C}_4\text{H}_6$ and a 4-carbon chain (butyne).
  • In compound I, the triple bond is between the first and second carbon atoms (But-1-yne).
  • In compound III, the triple bond is between the second and third carbon atoms (But-2-yne).
  • They differ only in the position of the triple bond.

Therefore, compounds I and III are positional isomers.
$\boxed{\text{I and III}}$